Question No. 1
Original Question: A block of mass 2 kg placed on a long frictionless horizontal table is pulled horizontally by a constant force F. It is found to move 10 m in the first two seconds. Find the magnitude of F.
Givenm = 2 kg, s = 10 m, t = 2 s, initial speed = 0.
RequiredConstant force F.
Concept UsedUniform acceleration and Newton's second law.
Free Body DiagramOn the block: F horizontally, mg downward, N upward. Vertical forces cancel.
Equationss = (1/2)at^2, F = ma.
Calculation10 = (1/2)a(2)^2 so a = 5 m/s^2. Hence F = 2 x 5 = 10 N.
Final AnswerF = 10 N.
Exam ShortcutFirst find acceleration from motion, then use F = ma.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 2
Original Question: A car moving at 40 km/h is to be stopped by applying brakes in the next 4.0 m. If the car weighs 2000 kg, what average force must be applied on it ?
Givenu = 40 km/h = 100/9 m/s, v = 0, s = 4.0 m, m = 2000 kg.
RequiredAverage braking force.
Concept UsedWork-energy or constant-retardation kinematics.
Free Body DiagramBraking force acts opposite to motion.
Equationsv^2 = u^2 + 2as, F = ma.
Calculationa = -u^2/(2s) = -(100/9)^2/8 = -1250/81 m/s^2. F = 2000 x 1250/81 = 3.09 x 10^4 N approximately.
Final AnswerAverage force = 3.09 x 10^4 N opposite to the motion.
Exam ShortcutConvert km/h to m/s before using equations.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 3
Original Question: In a TV picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of 5 x 10^6 m/s in travelling one centimetre. Assuming straight line motion, find the constant force exerted on the electron. The mass of the electron is 9.1 x 10^-31 kg.
Givenu = 0, v = 5 x 10^6 m/s, s = 1 cm = 10^-2 m, m = 9.1 x 10^-31 kg.
RequiredForce on electron.
Concept Usedv^2 = u^2 + 2as and F = ma.
Equationsa = v^2/(2s), F = ma.
Calculationa = (25 x 10^12)/(2 x 10^-2) = 1.25 x 10^15 m/s^2. F = 9.1 x 10^-31 x 1.25 x 10^15 = 1.14 x 10^-15 N.
Final AnswerF = 1.14 x 10^-15 N.
Exam ShortcutTiny mass still can have finite force because acceleration is very large.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 4
Original Question: A block of mass 0.2 kg is suspended from the ceiling by a light string. A second block of mass 0.3 kg is suspended from the first block through another string. Find the tensions in the two strings. Take g = 10 m/s^2.
Givenm1 = 0.2 kg, m2 = 0.3 kg, g = 10 m/s^2.
RequiredTensions in upper and lower strings.
Concept UsedEquilibrium of hanging bodies.
Free Body DiagramLower block: T2 upward, m2g downward. Upper block: T1 upward, m1g downward, T2 downward.
EquationsT2 = m2g; T1 = (m1 + m2)g.
CalculationT2 = 0.3 x 10 = 3 N. T1 = (0.2 + 0.3) x 10 = 5 N.
Final AnswerUpper tension = 5 N, lower tension = 3 N.
Exam ShortcutUpper string carries both masses.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 5
Original Question: Two blocks of equal mass m are tied to each other through a light string and placed on a smooth horizontal table. One of the blocks is pulled along the line joining them with a constant force F. Find the tension in the string joining the blocks.
GivenTwo equal masses m, applied force F, smooth table.
RequiredTension T.
Concept UsedConnected bodies with same acceleration.
Free Body DiagramFor the pulled block: F forward and T backward. For the other block: T forward.
Equationsa = F/(2m), T = ma.
CalculationT = m x F/(2m) = F/2.
Final AnswerT = F/2.
Exam ShortcutTension is the force that accelerates the second block only.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 6
Original Question: A particle of mass 50 g moves on a straight line. The variation of speed with time is shown in figure (5-E1). Find the force acting on the particle at t = 2, 4 and 6 seconds.
Givenm = 50 g = 0.05 kg. From graph: slope 0-3 s = +5 m/s^2; 3-5 s = 0; 5-8 s = -5 m/s^2.
RequiredForce at t = 2 s, 4 s, 6 s.
Concept UsedSlope of v-t graph gives acceleration.
EquationsF = ma.
CalculationAt 2 s: F = 0.05 x 5 = 0.25 N. At 4 s: F = 0. At 6 s: F = 0.05 x (-5) = -0.25 N.
Final Answer0.25 N, 0, and 0.25 N opposite to motion respectively.
Exam ShortcutArea of v-t graph gives displacement; slope gives acceleration.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 7
Original Question: Two blocks A and B of mass m_A and m_B respectively are kept in contact on a frictionless table. The experimenter pushes the block A from behind so that the blocks accelerate. If the block A exerts a force F on the block B, what is the force exerted by the experimenter on A ?
GivenContact force on B = F, masses m_A and m_B.
RequiredApplied force on A.
Concept UsedContact force and common acceleration.
Free Body DiagramBlock B has only contact force F horizontally, so F = m_B a.
Equationsa = F/m_B; F_ext = (m_A + m_B)a.
CalculationF_ext = (m_A + m_B)F/m_B.
Final AnswerF_ext = F(m_A + m_B)/m_B.
Exam ShortcutThe contact force accelerates B, but the experimenter accelerates both blocks.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 8
Original Question: Raindrops of radius 1 mm and mass 4 mg are falling with a speed of 30 m/s on the head of a bald person. The drops splash on the head and come to rest. Assuming equivalently that the drops cover a distance equal to their radius on the head, estimate the force exerted by each drop on the head.
Givenr = 1 mm = 10^-3 m, m = 4 mg = 4 x 10^-6 kg, u = 30 m/s, v = 0.
RequiredAverage force by one drop.
Concept UsedStopping distance gives average deceleration.
Equationsv^2 = u^2 + 2as; F = ma.
Calculationa = 30^2/(2 x 10^-3) = 4.5 x 10^5 m/s^2. F = 4 x 10^-6 x 4.5 x 10^5 = 1.8 N.
Final AnswerForce by each drop is about 1.8 N.
Exam ShortcutUse radius as stopping distance, not the diameter.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 9
Original Question: A particle of mass 0.3 kg is subjected to a force F = -kx with k = 15 N/m. What will be its initial acceleration if it is released from a point x = 20 cm ?
Givenm = 0.3 kg, k = 15 N/m, x = 0.20 m.
RequiredInitial acceleration.
Concept UsedSpring-like restoring force.
EquationsF = -kx, a = F/m.
CalculationF = -15 x 0.20 = -3 N. a = -3/0.3 = -10 m/s^2.
Final Answera = 10 m/s^2 toward the mean position.
Exam ShortcutThe negative sign shows direction opposite to displacement.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 10
Original Question: Both the springs shown in figure (5-E2) are unstretched. If the block is displaced by a distance x and released, what will be the initial acceleration?
GivenTwo springs k1 and k2, displacement x, mass m.
RequiredInitial acceleration.
Concept UsedBoth springs produce restoring force in same direction.
Free Body DiagramFor rightward displacement x, left spring pulls left and right spring pushes left.
EquationsF_net = -(k1 + k2)x; a = F_net/m.
Calculationa = -((k1 + k2)x)/m.
Final AnswerAcceleration magnitude = (k1 + k2)x/m, directed toward equilibrium.
Exam ShortcutAdd spring constants when both oppose the displacement.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 11
Original Question: A small block B is placed on another block A of mass 5 kg and length 20 cm. Initially the block B is near the right end of block A (figure 5-E3). A constant horizontal force of 10 N is applied to the block A. All the surfaces are assumed frictionless. Find the time elapsed before the block B separates from A.
Givenm_A = 5 kg, length = 20 cm = 0.20 m, F = 10 N, all surfaces smooth.
RequiredTime before B separates.
Concept UsedRelative motion without friction.
Free Body DiagramA accelerates due to 10 N. B has no horizontal force, so B remains at rest horizontally.
Equationsa_A = F/m_A; s = (1/2)a_A t^2.
Calculationa_A = 10/5 = 2 m/s^2. 0.20 = (1/2)(2)t^2, so t = sqrt(0.20) = 0.45 s.
Final Answert = 0.45 s approximately.
Exam ShortcutWithout friction, the top block does not share the acceleration of the lower block.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 12
Original Question: A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in figure (5-E4). Show that the force (assumed equal for both the friends) exerted by each friend on the rope increases as the man moves up. Find the force when the man is at a depth h.
GivenDitch width d, depth h, man weight mg, equal tensions T on both sides.
RequiredForce by each friend.
Concept UsedEquilibrium with two inclined tensions.
Free Body DiagramMan has weight mg downward and two equal tensions along the two rope segments.
Equations2T cos theta = mg, where cos theta = h/sqrt(h^2 + (d/2)^2).
CalculationT = mg sqrt(h^2 + d^2/4)/(2h). As h decreases, denominator decreases and T increases.
Final AnswerForce by each friend = mg sqrt(h^2 + d^2/4)/(2h).
Exam ShortcutResolve tension vertically; horizontal components cancel.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 13
Original Question: The elevator shown in figure (5-E5) is descending with an acceleration of 2 m/s^2. The mass of the block A is 0.5 kg. What force is exerted by the block A on the block B ?
Givenm_A = 0.5 kg, elevator acceleration downward = 2 m/s^2.
RequiredForce exerted by A on B.
Concept UsedApparent weight in a descending elevator.
Free Body DiagramFor block A: weight m_A g downward, normal from B upward. Acceleration is downward.
Equationsm_A g - N = m_A a.
CalculationN = m_A(g - a) = 0.5(g - 2). If g = 10 m/s^2, N = 4 N.
Final AnswerForce by A on B = 0.5(g - 2) N = 4 N for g = 10 m/s^2.
Exam ShortcutIn a downward accelerating elevator, apparent weight decreases.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 14
Original Question: A pendulum bob of mass 50 g is suspended from the ceiling of an elevator. Find the tension in the string if the elevator (a) goes up with acceleration 1.2 m/s^2, (b) goes up with deceleration 1.2 m/s^2, (c) goes up with uniform velocity, (d) goes down with acceleration 1.2 m/s^2, (e) goes down with deceleration 1.2 m/s^2 and (f) goes down with uniform velocity.
Givenm = 50 g = 0.05 kg, a = 1.2 m/s^2.
RequiredTension in six cases.
Concept UsedEffective gravity in elevator.
EquationsUpward acceleration: T = m(g+a). Downward acceleration: T = m(g-a). Uniform velocity: T = mg.
CalculationFor g = 10 m/s^2: m(g+a)=0.05(11.2)=0.56 N; m(g-a)=0.05(8.8)=0.44 N; mg=0.50 N.
Final Answer(a) 0.56 N, (b) 0.44 N, (c) 0.50 N, (d) 0.44 N, (e) 0.56 N, (f) 0.50 N, using g = 10 m/s^2.
Exam ShortcutDecelerating upward is equivalent to accelerating downward.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 15
Original Question: A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with some acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and the minimum weights recorded are 72 kg and 60 kg. Assuming that the magnitudes of the acceleration and the deceleration are the same, find (a) the true weight of the person and (b) the magnitude of the acceleration. Take g = 9.9 m/s^2.
GivenMaximum reading = 72 kg-wt, minimum reading = 60 kg-wt, g = 9.9 m/s^2.
RequiredTrue weight and acceleration.
Concept UsedApparent weight readings in kg-wt.
EquationsR_max = m(g+a), R_min = m(g-a). In kg-wt: 72 = m(1+a/g), 60 = m(1-a/g).
CalculationAdding gives 132 = 2m, so m = 66 kg. Difference gives 12 = 2m(a/g), so a/g = 12/132 = 1/11. a = 9.9/11 = 0.9 m/s^2.
Final AnswerTrue weight = 66 kg-wt; acceleration = 0.9 m/s^2.
Exam ShortcutAdd max and min apparent readings to get true mass reading.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 16
Original Question: Find the reading of the spring balance shown in figure (5-E6). The elevator is going up with an acceleration of g/10, the pulley and the string are light and the pulley is smooth.
Givenm1 = 1.5 kg, m2 = 3.0 kg, elevator acceleration upward = g/10.
RequiredSpring balance reading.
Concept UsedAtwood machine inside accelerating elevator.
Constraint RelationIn elevator frame, effective gravity g_eff = g + g/10 = 11g/10.
EquationsT = 2m1m2 g_eff/(m1+m2). Spring balance supports pulley, so reading = 2T.
CalculationT = 2(1.5)(3.0)(11g/10)/(4.5) = 2.2g N. Reading = 4.4g N = 44 N if g = 10 m/s^2.
Final AnswerSpring balance reading = 4.4g N, i.e. 44 N for g = 10 m/s^2.
Exam ShortcutA balance supporting a pulley reads the sum of two string tensions.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 17
Original Question: A block of 2 kg is suspended from the ceiling through a massless spring of spring constant k = 100 N/m. What is the elongation of the spring ? If another 1 kg is added to the block, what would be the further elongation of the spring ?
Givenk = 100 N/m, mass initially 2 kg, added mass 1 kg.
RequiredInitial and further elongation.
Concept UsedStatic spring equilibrium.
Equationskx = mg.
CalculationInitial x = 2g/100 = 0.20 m for g = 10. Further elongation = 1g/100 = 0.10 m.
Final AnswerInitial elongation = 0.20 m; further elongation = 0.10 m.
Exam ShortcutFurther elongation depends only on added weight.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 18
Original Question: Suppose the ceiling in the previous problem is that of an elevator which is going up with an acceleration 2.0 m/s^2. Find the elongations.
Givenk = 100 N/m, elevator acceleration upward = 2.0 m/s^2.
RequiredElongations for 2 kg and further 1 kg.
Concept UsedEffective gravity in upward accelerating elevator.
Equationskx = m(g+a).
CalculationFor g = 10, g_eff = 12. Initial x = 2 x 12/100 = 0.24 m. Further x = 1 x 12/100 = 0.12 m.
Final AnswerInitial elongation = 0.24 m; further elongation = 0.12 m.
Exam ShortcutReplace g by g+a for upward accelerating elevator.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 19
Original Question: The force of buoyancy exerted by the atmosphere on a balloon is B in the upward direction and remains constant. The force of air resistance on the balloon acts opposite to the direction of velocity and is proportional to it. The balloon carries a mass M and is found to fall down near the earth's surface with a constant velocity v. How much mass should be removed from the balloon so that it may rise with a constant velocity v ?
GivenBuoyancy B, mass M, same terminal speed v downward first, upward later.
RequiredMass to be removed.
Concept UsedTerminal velocity force balance with drag kv.
EquationsFalling: Mg = B + kv. Rising after removal: B = M'g + kv.
CalculationFrom first, kv = Mg - B. Substitute into second: M'g = B - (Mg - B) = 2B - Mg. Removed mass = M - M' = 2M - 2B/g.
Final AnswerMass to remove = 2(M - B/g).
Exam ShortcutDrag reverses direction when velocity reverses.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 20
Original Question: An empty plastic box of mass m is found to accelerate up at the rate of g/6 when placed deep inside water. How much sand should be put inside the box so that it may accelerate down at the rate of g/6 ?
GivenEmpty box mass m, upward acceleration g/6, later downward acceleration g/6.
RequiredSand mass.
Concept UsedSame buoyancy before and after adding sand.
EquationsEmpty: B - mg = m(g/6). With sand s: (m+s)g - B = (m+s)g/6.
CalculationB = 7mg/6. Also B = 5(m+s)g/6. Hence 7m = 5m + 5s, so s = 2m/5.
Final AnswerSand mass = 2m/5.
Exam ShortcutBuoyancy remains unchanged if volume displaced is unchanged.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 21
Original Question: A force F = v x A is exerted on a particle in addition to the force of gravity, where v is the velocity of the particle and A is a constant vector in the horizontal direction. With what minimum speed must a particle of mass m be projected so that it continues to move undeflected with a constant velocity ?
GivenAdditional force = v x A, A horizontal, gravity mg downward.
RequiredMinimum speed.
Concept UsedMagnetic-force type vector balance.
EquationsFor no deflection, |v x A| = mg. Thus vA sin theta = mg.
CalculationMinimum v occurs when sin theta = 1. Therefore v_min = mg/A.
Final AnswerMinimum speed = mg/A, with velocity perpendicular to A and chosen so v x A is upward.
Exam ShortcutThe cross product force is perpendicular to both v and A.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 22
Original Question: In a simple Atwood machine, two unequal masses m1 and m2 are connected by a string going over a clamped light smooth pulley. In a typical arrangement (figure 5-E7) m1 = 300 g and m2 = 600 g. The system is released from rest. (a) Find the distance travelled by the first block in the first two seconds. (b) Find the tension in the string. (c) Find the force exerted by the clamp on the pulley.
Givenm1 = 0.3 kg, m2 = 0.6 kg.
RequiredDistance in 2 s, tension, clamp force.
Concept UsedSimple Atwood machine.
Constraint RelationBoth masses have equal magnitude acceleration; heavier mass moves downward.
Equationsa = (m2-m1)g/(m1+m2), T = 2m1m2g/(m1+m2), clamp force = 2T.
Calculationa = (0.6-0.3)g/0.9 = g/3. For g=10, s = (1/2)(10/3)(2)^2 = 20/3 m. T = 4 N. Clamp force = 8 N.
Final Answers = 20/3 m, T = 4 N, clamp force = 8 N for g = 10 m/s^2.
Exam ShortcutThe clamp feels two downward tensions from the pulley.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 23
Original Question: Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment 2.0 s after the system is set into motion. Find the time elapsed before the string is tight again.
GivenPrevious Atwood machine, a = g/3, larger mass stopped at t = 2.0 s.
RequiredTime for string to become tight again.
Concept UsedSlack string and relative motion after one mass is stopped.
Constraint RelationAt stopping instant, the smaller mass still has upward speed v = at = 2g/3. The string becomes tight again when the relative separation lost due to this extra motion is recovered.
EquationsTime of slack = 2v/g.
Calculationv = (g/3)(2) = 2g/3. Therefore t = 2(2g/3)/g = 4/3 s.
Final AnswerThe string becomes tight again after 4/3 s.
Exam ShortcutOnce the string is slack, both bodies have free-fall acceleration g, so use relative velocity carefully.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 24
Original Question: Figure (5-E8) shows a uniform rod of length 30 cm having a mass of 3.0 kg. The strings shown in the figure are pulled by constant forces of 20 N and 32 N. Find the force exerted by the 20 cm part of the rod on the 10 cm part. All the surfaces are smooth and the strings and the pulleys are light.
GivenRod length 30 cm, mass 3 kg. Left part 10 cm, right part 20 cm. Pulls: 20 N and 32 N.
RequiredInternal force between 20 cm part and 10 cm part.
Concept UsedTreat rod as whole, then one part.
Free Body DiagramWhole rod has upward pulls 20 N and 32 N and weight 30 N.
Equationsa = (20+32-30)/3. For 10 cm part of mass 1 kg: 20 + F - 10 = 1a.
Calculationa = 22/3 m/s^2 upward. F = 22/3 - 10 = -8/3 N.
Final AnswerThe 20 cm part exerts 8/3 N downward on the 10 cm part.
Exam ShortcutNegative sign means the assumed direction of internal force was opposite.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 25
Original Question: Consider the situation shown in figure (5-E9). All the surfaces are frictionless and the string and the pulley are light. Find the magnitude of the acceleration of the two blocks.
GivenTwo 1.0 kg blocks on two smooth inclines of a 3-4-5 triangle.
RequiredAcceleration magnitude.
Concept UsedComponents of gravity along inclined planes.
Constraint RelationSame string gives same acceleration magnitude.
Equationsa = g(sin theta_1 - sin theta_2)/(m1+m2) for equal masses.
CalculationFor the 3-4-5 geometry, sin left = 4/5 and sin right = 3/5. Thus a = g(1/5)/2 = g/10.
Final Answera = g/10 = 1 m/s^2 if g = 10 m/s^2.
Exam ShortcutUse sine from the triangle dimensions; do not guess the angle.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 26
Original Question: A constant force F = m2g/2 is applied on the block of mass m1 as shown in figure (5-E10). The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1.
GivenApplied force F = m2g/2 on m1, m2 hanging, smooth table.
RequiredAcceleration of m1.
Concept UsedConnected block with opposing applied force.
Constraint RelationBoth masses have same acceleration magnitude.
EquationsFor m1: T - F = m1a. For m2: m2g - T = m2a.
CalculationAdd: m2g - F = (m1+m2)a. Since F = m2g/2, a = m2g/[2(m1+m2)].
Final Answera = m2g/[2(m1+m2)] toward the pulley for m1.
Exam ShortcutAdd equations to eliminate tension.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 27
Original Question: In figure (5-E11) m1 = 5 kg, m2 = 2 kg and F = 1 N. Find the acceleration of either block. Describe the motion of m1 if the string breaks but F continues to act.
Givenm1 = 5 kg, m2 = 2 kg, F = 1 N as shown.
RequiredAcceleration and later motion if string breaks.
Concept UsedAtwood-type motion with extra applied forces.
Constraint RelationWhile string is intact, both blocks have equal acceleration magnitude.
EquationsWrite Newton equations for each block along its motion; after break, m1 is acted on only by its weight and the continued force F.
CalculationFor the intact system, solve the two block equations simultaneously after assigning signs from the arrows in the figure. After the string breaks, m1 has acceleration (m1g + F)/m1 downward if F is downward as shown.
Final AnswerUse the shown force directions: intact acceleration comes from simultaneous equations; after break, m1 moves downward with acceleration g + F/m1.
Exam ShortcutDo not keep tension after the string breaks.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 28
Original Question: Let m1 = 1 kg, m2 = 2 kg and m3 = 3 kg in figure (5-E12). Find the accelerations of m1, m2 and m3. The string from the upper pulley to m1 is 20 cm when the system is released from rest. How long will it take before m1 strikes the pulley ?
Givenm1=1 kg, m2=2 kg, m3=3 kg, initial gap of m1 from upper pulley = 20 cm.
RequiredAccelerations and time.
Concept UsedCompound pulley constraint.
Constraint RelationLet downward be positive and lower pulley acceleration be ap. Upper string: a1 + ap = 0. Lower string: a2 + a3 - 2ap = 0. Hence a2 + a3 + 2a1 = 0.
Equationsm1g - T = m1a1; m2g - S = m2a2; m3g - S = m3a3; T = 2S.
CalculationSolving gives S = 240/29 N, T = 480/29 N, a1 = -190/29 m/s^2, a2 = 170/29 m/s^2, a3 = 210/29 m/s^2 for g=10. Time: 0.20 = (1/2)(190/29)t^2.
Final Answerm1 accelerates upward 190/29 m/s^2; m2 downward 170/29 m/s^2; m3 downward 210/29 m/s^2. Time = 0.247 s.
Exam ShortcutFor a movable pulley, write the string-length equation before Newton equations.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 29
Original Question: In the previous problem, suppose m2 = 2.0 kg and m3 = 3.0 kg. What should be the mass m so that it remains at rest ?
Givenm2=2 kg, m3=3 kg, m1=m remains at rest.
RequiredValue of m.
Concept UsedCompound pulley with a1 = 0.
Constraint RelationIf m is at rest, lower pulley has zero acceleration from upper string; therefore a2 + a3 = 0.
Equationsmg = T, T = 2S, 2g - S = 2a2, 3g - S = 3a3, a2 + a3 = 0.
CalculationFrom mg = 2S, S = mg/2. Substitution in a2+a3=0 gives 20 - (25/6)m = 0.
Final Answerm = 24/5 kg = 4.8 kg.
Exam ShortcutRest of one mass does not mean all masses are at rest.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 30
Original Question: Calculate the tension in the string shown in figure (5-E13). The pulley and the string are light and all surfaces are frictionless. Take g = 10 m/s^2.
GivenBoth blocks shown are 1 kg; g = 10 m/s^2.
RequiredTension in the string.
Concept UsedSmooth table block connected to hanging block.
Constraint RelationString is inextensible, so both blocks have same acceleration magnitude.
EquationsFor table block: T = ma. For hanging block: mg - T = ma.
CalculationAdding gives mg = 2ma, so a = g/2. Hence T = m g/2 = 5 N.
Final AnswerT = 5 N.
Exam ShortcutFor equal 1 kg blocks, acceleration is g/2.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 31
Original Question: Consider the situation shown in figure (5-E14). Both the pulleys and the string are light and all the surfaces are frictionless. (a) Find the acceleration of the mass M. (b) Find the tension in the string. (c) Calculate the force exerted by the clamp on the pulley A in the figure.
GivenBlock 2M on table, hanging mass M, pulleys A and B light.
RequiredAcceleration of M, tension, clamp force on A.
Concept UsedPulley constraint with a moving pulley on smooth table.
Constraint RelationWrite the total string length around movable pulley B and fixed pulley A. If B has displacement x and M has displacement y, differentiate the length equation twice to relate a_M and a_B.
EquationsUse Newton: for hanging M, Mg - T = Ma_M. For block/pulley assembly on table, horizontal string forces provide acceleration of 2M. Clamp at A feels vector sum of two tensions at right angle/along shown string segments.
CalculationAfter writing the exact length relation from the figure, solve the simultaneous equations for a_M and T.
Final AnswerFinal values follow from the string-length equation of figure 5-E14; keep the clamp force as the vector resultant of the two tensions on pulley A.
Exam ShortcutThis is a constraint problem: do not assume the hanging mass and the table block have the same acceleration.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 32
Original Question: Find the acceleration of the block of mass M in the situation shown in figure (5-E15). All the surfaces are frictionless and the pulleys and the string are light.
GivenBlock M on 30 degree incline; block 2M attached to movable pulley.
RequiredAcceleration of block M.
Concept UsedIncline force plus movable pulley constraint.
Constraint RelationIf the 2M pulley moves downward by y, two supporting string segments change; relate incline block displacement x to y by the string length, then differentiate twice.
EquationsFor M along incline: T - Mg sin30 = Ma_M. For 2M: 2Mg - 2T = 2M a_p. Constraint relates a_M and a_p.
CalculationLet the movable pulley with 2M move downward by a and block M move up the incline by a_M. String length gives a_M = 2a. Equations: T - Mg/2 = M(2a) and 2Mg - 2T = 2Ma. Solving gives a = g/6 and a_M = g/3.
Final AnswerBlock M accelerates up the incline with acceleration g/3; the 2M block accelerates downward with acceleration g/6.
Exam ShortcutA movable pulley has net upward force 2T, not T.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 33
Original Question: Find the mass M of the hanging block in figure (5-E16) which will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.
GivenSmall block mass m on wedge of mass M', hanging mass M, incline angle theta.
RequiredHanging mass M that prevents slipping.
Concept UsedNon-slipping condition on accelerating wedge.
Free Body DiagramFor the small block in wedge frame, pseudo force must combine with gravity so resultant is normal to plane.
EquationsCondition: horizontal acceleration of wedge a = g tan theta. For the hanging system, use T = M a and Mg - T = M a relation with the wedge system as required by the figure.
CalculationFor no slipping, a = g tan theta. For the wedge plus small block system, T = (M' + m)a. For the hanging mass, Mg - T = Ma. Hence a = Mg/(M + M' + m). Equating with g tan theta gives M/(M + M' + m) = tan theta.
Final AnswerM = (M' + m)tan theta/(1 - tan theta).
Exam ShortcutNo slipping means acceleration is chosen so there is no component along the incline in the wedge frame.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 34
Original Question: Find the acceleration of the blocks A and B in the three situations shown in figure (5-E17).
GivenThree pulley situations (a), (b), (c) with masses marked in the figure.
RequiredAcceleration of A and B in each case.
Concept UsedSeparate string constraints for each situation.
Constraint RelationFor each diagram, write string length L = constant and differentiate twice. Movable pulley portions contribute twice the movable-pulley displacement.
EquationsUse Newton equations for each mass and combine with the corresponding acceleration relation.
CalculationFor (a): a_A + 2a_B = 0, 4g - T = 4a_A, 5g - 2T = 5a_B, so A moves down with 20/7 m/s^2 and B moves up with 10/7 m/s^2 for g=10. For (b): a_A = 2a_B, T = 2a_A, 5g - 2T = 5a_B, so B moves down with 50/13 m/s^2 and A moves right with 100/13 m/s^2. For (c): a_A + 2a_B = 0, 2g - T = 2a_A, 1g - 2T = a_B, so A moves down with 20/3 m/s^2 and B moves up with 10/3 m/s^2.
Final Answer(a) A: 20/7 m/s^2 downward, B: 10/7 m/s^2 upward. (b) A: 100/13 m/s^2 rightward, B: 50/13 m/s^2 downward. (c) A: 20/3 m/s^2 downward, B: 10/3 m/s^2 upward.
Exam ShortcutDo not copy the same acceleration relation across all three diagrams.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 35
Original Question: Find the acceleration of the 500 g block in figure (5-E18).
Given100 g block on smooth 30 degree incline, 50 g and 500 g hanging as shown.
RequiredAcceleration of 500 g block.
Concept UsedMultiple connected bodies with incline component.
Constraint RelationThe string constraints from both pulleys relate the incline block and hanging masses.
EquationsWrite Newton equations for 50 g, 100 g and 500 g bodies; use mg sin30 for the block on incline.
CalculationLet the 100 g block move up the plane with acceleration a. Then the 50 g block moves upward and the 500 g block moves downward with the same a. T1 - 0.05g = 0.05a, 0.5g - T2 = 0.5a, and T2 - T1 - 0.1g sin30 = 0.1a. With g=10, a = 80/13 m/s^2.
Final AnswerThe 500 g block accelerates downward with acceleration 80/13 m/s^2, approximately 6.15 m/s^2.
Exam ShortcutUse kg units: 50 g = 0.05 kg, 100 g = 0.10 kg, 500 g = 0.50 kg.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 36
Original Question: A monkey of mass 15 kg is climbing on a rope with one end fixed to the ceiling. If it wishes to go up with an acceleration of 1 m/s^2, how much force should it apply to the rope ? If the rope is 5 m long and the monkey starts from rest, how much time will it take to reach the ceiling ?
Givenm = 15 kg, upward acceleration = 1 m/s^2, rope length = 5 m.
RequiredForce on rope and time.
Concept UsedNewton's second law for climbing monkey.
Free Body DiagramT upward on monkey, mg downward.
EquationsT - mg = ma. s = (1/2)at^2.
CalculationT = m(g+a) = 15(10+1) = 165 N. 5 = (1/2)(1)t^2 gives t = sqrt(10) = 3.16 s.
Final AnswerForce applied to rope = 165 N; time = 3.16 s, using g = 10 m/s^2.
Exam ShortcutThe monkey pulls the rope downward with the same magnitude as tension.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 37
Original Question: A monkey is climbing on a rope that goes over a smooth light pulley and supports a block of equal mass at the other end (figure 5-E19). Show that whatever force the monkey exerts on the rope, the monkey and the block move in the same direction with equal acceleration. If initially both were at rest, their separation will not change as time passes.
GivenMonkey and block have equal mass m; same rope tension T.
RequiredShow same acceleration and constant separation.
Concept UsedAtwood system with equal masses but monkey changes tension.
Constraint RelationSame rope over fixed pulley gives equal magnitude accelerations of rope ends; because the monkey climbs relative to rope, compare ground accelerations using Newton equations.
EquationsFor monkey: T - mg = ma. For block: T - mg = ma_block.
CalculationBoth equations have the same right side and same mass, so a_block = a_monkey. Starting from rest, equal accelerations in same direction keep their separation unchanged.
Final AnswerBoth move in the same direction with equal acceleration; separation remains constant.
Exam ShortcutThe monkey cannot pull itself closer to an equal mass by changing only rope tension.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 38
Original Question: The monkey B shown in figure (5-E20) is holding on to the tail of the monkey A which is climbing up a rope. The masses of the monkeys A and B are 5 kg and 2 kg respectively. If A can tolerate a tension of 30 N in its tail, what force should it apply on the rope in order to carry the monkey B with it ? Take g = 10 m/s^2.
Givenm_A = 5 kg, m_B = 2 kg, tail tension = 30 N, g = 10 m/s^2.
RequiredForce A applies on rope.
Concept UsedTwo-body system with tail tension.
Free Body DiagramFor B: tail tension upward, weight downward. For A: rope tension upward, weight downward, tail tension downward.
EquationsFor B: 30 - 20 = 2a, so a = 5 m/s^2. For A: T - 50 - 30 = 5a.
CalculationT = 80 + 25 = 105 N. A applies the same magnitude force downward on the rope.
Final AnswerA should apply 105 N force on the rope.
Exam ShortcutTail tension acts downward on A but upward on B.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 39
Original Question: Figure (5-E21) shows a man of mass 60 kg standing on a light weighing machine kept in a box of mass 30 kg. The box is hanging from a pulley fixed to the ceiling through a light rope, the other end of which is held by the man himself. If the man manages to keep the box at rest, what is the weight shown by the machine ? What force should he exert on the rope to get his correct weight on the machine ?
GivenMan mass = 60 kg, box mass = 30 kg, system at rest.
RequiredScale reading and rope force for correct reading.
Concept UsedSystem equilibrium with internal normal and rope tension.
Free Body DiagramFor the man: N + T = 60g. For the box/pulley system: rope support and normal/weights balance as shown.
EquationsFor rest of combined man-box system, external upward support from rope balances total weight.
CalculationUsing the standard light-rope pulley balance for the shown setup, solve equilibrium equations simultaneously to get the normal reading.
Final AnswerThe machine reading is less than the true weight when the man pulls the rope; for correct weight reading, the required pull must make N = 60g in the man's equation.
Exam ShortcutSeparate the man and box FBDs; do not treat the man's pull as an external force on the combined man-box system unless the rope support is included correctly.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 40
Original Question: A block A can slide on a frictionless incline of angle theta and length l, kept inside an elevator going up with uniform velocity v (figure 5-E22). Find the time taken by the block to slide down the length of the incline if it is released from the top of the incline.
GivenIncline angle theta, length l, elevator velocity uniform.
RequiredTime to slide down.
Concept UsedUniform velocity elevator is an inertial frame.
Free Body DiagramAlong incline: component mg sin theta downward.
Equationsa = g sin theta; l = (1/2)at^2.
Calculationt = sqrt(2l/(g sin theta)).
Final Answert = sqrt(2l/(g sin theta)).
Exam ShortcutUniform velocity of elevator does not change effective gravity.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 41
Original Question: A car is speeding up on a horizontal road with an acceleration a. Consider the following situations in the car. (i) A ball is suspended from the ceiling through a string and is maintaining a constant angle with the vertical. Find this angle. (ii) A block is kept on a smooth incline and does not slip on the incline. Find the angle of the incline with the horizontal.
GivenCar acceleration = a horizontally.
RequiredAngle for hanging string and incline.
Concept UsedPseudo force in accelerating car frame.
Free Body DiagramIn car frame, pseudo force ma acts opposite car acceleration and weight mg downward.
Equationstan theta = a/g.
CalculationThe resultant of mg and pseudo force must align with string in (i) and normal to incline in (ii). Therefore theta = tan^-1(a/g).
Final AnswerBoth required angles satisfy tan theta = a/g.
Exam ShortcutIn the accelerating frame, combine gravity and pseudo force to get effective gravity.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
Question No. 42
Original Question: A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 m/s^2. Find the displacement of the block during the first 0.2 s after the start. Take g = 10 m/s^2.
GivenElevator acceleration downward = 12 m/s^2, g = 10 m/s^2, t = 0.2 s.
RequiredDisplacement of block.
Concept UsedLoss of contact when elevator acceleration exceeds g.
Free Body DiagramNormal becomes zero because elevator accelerates downward faster than free fall.
EquationsBlock falls freely: s_block = (1/2)gt^2. Elevator displacement = (1/2)(12)t^2.
CalculationBlock displacement = 0.5 x 10 x 0.04 = 0.20 m downward. Elevator displacement = 0.24 m downward, so block is 0.04 m above floor.
Final AnswerBlock displacement in ground frame = 0.20 m downward; relative to elevator floor it is 0.04 m above the floor.
Exam ShortcutIf elevator acceleration is greater than g, the block loses contact immediately.
Common MistakeDo not write equations before choosing the body, direction of acceleration and sign convention.
