CBSE Class 11 Physics Newton laws of Motion Objective Questions

OBJECTIVES
CLASS 12 PHYSICS : NEWTON'S LAWS OF MOTION

1. A boy of mass 60 kg is holding a vertical rope. The rope can break when a mass of 75 kg is suspended from it. The maximum acceleration with which the boy can climb the rope without breaking it is
(a) 2.5 m/s²
(b) 5.0 m/s²
(c) 7.5 m/s²
(d) 22.5 m/s²

Answer: (a) 2.5 m/s²

1. Take g = 10 m/s². The maximum tension the rope can bear = 75g = 750 N.
2. While climbing up with acceleration a, the tension is T = m(g + a) = 60(10 + a).
3. 60(10 + a) = 750, so 10 + a = 12.5 and a = 2.5 m/s².
4. Hence, the correct option is (a).

Teacher's Note:
a) The rope tension is greater than the weight when the boy accelerates upward.
b) The breaking load of 75 kg means a force of 75g.

2. A pendulum with a bob of 0.1 kg is suspended from the roof of a car which is moving with an acceleration of 5 m/s² relative to the ground towards right. An observer travelling in the car experiences a pseudo force acting on the bob equal to
(a) 1.0 Newton downwards
(b) 1.0 Newton upwards
(c) 0.5 Newton towards right
(d) 0.5 Newton towards left

Answer: (d) 0.5 Newton towards left

1. The pseudo force on a body in an accelerating frame is F = -ma, directed opposite to the acceleration of the frame.
2. F = 0.1 × 5 = 0.5 N.
3. The car accelerates to the right, so the pseudo force is 0.5 N towards the left.
4. Hence, the correct option is (d).

Teacher's Note:
a) The pseudo force is only for the observer inside the accelerating car.
b) Its direction is always opposite to the acceleration of the frame.

3. The mass of an elevator is 4000 kg. Starting from rest how much distance will it rise in 2 s, if the tension in the cable is 56 kN?
(a) 3 m
(b) 4 m
(c) 8 m
(d) 16 m

Answer: (c) 8 m

1. Take g = 10 m/s². Net upward force = 56000 - 4000 × 10 = 16000 N.
2. a = 16000/4000 = 4 m/s².
3. s = (1/2)at² = (1/2) × 4 × 2² = 8 m.
4. Hence, the correct option is (c).

Teacher's Note:
a) Subtract the weight from the cable tension to find the net force.
b) Convert kN to N first.

4. The weights W and 2W are suspended from the ends of a light string passing over a smooth fixed pulley. If the pulley is pulled up with an acceleration of 10 m/s², the tension in the string will be
(a) \( \frac{W}{3} \)
(b) \( \frac{2W}{3} \)
(c) \( \frac{4W}{3} \)
(d) \( \frac{8W}{3} \)

Answer: (d) \( \frac{8W}{3} \)

1. Take g = 10 m/s². In the frame of the pulley, which accelerates upward at 10 m/s², the effective acceleration due to gravity is g' = g + 10 = 20 m/s².
2. For an Atwood machine, T = 2m₁m₂g'/(m₁ + m₂) with m₁ = W/g and m₂ = 2W/g.
3. T = 2 × (W/g)(2W/g) × 20/(3W/g) = 80W/(3g) = 8W/3 (with g = 10).
4. Hence, the correct option is (d).

Teacher's Note:
a) Effective gravity is g + a when the support accelerates upward.
b) Without the pulley acceleration the tension would be 4W/3.

5. A pendulum having from the ceiling of a railway carriage makes an angle of 30º with the vertical when it is accelerating. The acceleration of the carriage in m/s² is
(a) \( \frac{10}{\sqrt{3}} \)
(b) \( \frac{20}{\sqrt{3}} \)
(c) 5√3
(d) 10√3

Answer: (a) \( \frac{10}{\sqrt{3}} \)

1. The bob is in equilibrium in the carriage frame under gravity, the tension and the pseudo force ma.
2. T sin 30° = ma and T cos 30° = mg, so a = g tan 30°.
3. a = 10 × (1/√3) = 10/√3 m/s².
4. Hence, the correct option is (a).

Teacher's Note:
a) The angle with the vertical gives tan θ = a/g.
b) Take g = 10 m/s² to match the options.

6. Two masses of 5 kg and 10 kg are suspended from a fixed support as shown in the figure. The system is pulled downwards with a force of 150 N applied to the lower mass. The string attached to the support breaks and the system accelerates downwards. If the downward force continues to act, the acceleration of the system is
(a) 5 m/s²
(b) 10 m/s²
(c) 20 m/s²
(d) 30 m/s²

Answer: (c) 20 m/s²

1. Take g = 10 m/s². After the support string breaks, the total downward force on the system = weight + applied force = 15 × 10 + 150 = 300 N.
2. Total mass = 5 + 10 = 15 kg.
3. a = 300/15 = 20 m/s².
4. Hence, the correct option is (c).

Teacher's Note:
a) Treat the two masses as one system, since the string between them stays taut.
b) The applied force of 150 N adds to the weight.

7. Two masses m and 3m are connected as shown in the figure. The tension in the string is
(a) \( \frac{3mg}{4} \)
(b) \( \frac{9mg}{8} \)
(c) mg
(d) 2mg

Answer: (b) \( \frac{9mg}{8} \)

1. The mass 3m on the 30° incline has the component of weight along the slope = 3mg sin 30° = 1.5mg, which is greater than the weight mg of the hanging mass.
2. So 3m moves down the slope and m moves up. a = (1.5mg - mg)/(3m + m) = g/8.
3. For the mass m: T - mg = m × g/8, so T = mg + mg/8 = 9mg/8.
4. Hence, the correct option is (b).

Teacher's Note:
a) The incline is taken as smooth. Compare the pulling forces on the two sides to find the direction of motion.
b) Check with the other mass: 3mg/2 - T = 3m × g/8 gives T = 9mg/8.

8. A block of mass 1.0 kg rests on a wedge of angle θ. The acceleration which should be given to the wedge so that the block falls freely is
(a) 10 m/s²
(b) 10 cos θ m/s²
(c) 10 cot θ m/s²
(d) 10 tan θ m/s²

Answer: (c) 10 cot θ m/s²

1. For the block to fall freely, the normal reaction on it must be zero, so the wedge surface must move away from the block just as fast as the block falls.
2. If the block falls a height y, the wedge must move horizontally a distance y cot θ in the same time (the slope has tan θ = y/x).
3. Equating the accelerations: a_wedge = g cot θ = 10 cot θ m/s².
4. Hence, the correct option is (c).

Teacher's Note:
a) A free fall means no force from the wedge on the block.
b) A steeper wedge needs a smaller horizontal acceleration.

9. Three equals weights A, B, C of mass 2 kg each are hanging on a string passing over a fixed frictionless pulley as shown in the figure. The tension in the string connecting weights B and C is
(a) 6.6 N
(b) 13.3 N
(c) 19.6 N
(d) 26.6 N

Answer: (b) 13.3 N

1. Take g = 10 m/s². Weight A (2 kg) hangs on one side; B and C together (4 kg) hang on the other side.
2. a = (4 - 2)g/(4 + 2) = g/3 = 3.33 m/s².
3. For C: mg - T = ma, so T = m(g - a) = 2(10 - 3.33) = 13.3 N.
4. Hence, the correct option is (b).

Teacher's Note:
a) The string between B and C carries only the weight of C (with its acceleration).
b) Treat B and C as a single 4 kg mass to find the acceleration.

10. Two masses of 10 kg and 20 kg are connected with an inextensible string passing over two smooth pulleys as shown in the figure. The acceleration of 20 kg mass will be
(a) \( \frac{10}{3} \) m/s²
(b) \( \frac{20}{3} \) m/s²
(c) \( \frac{40}{9} \) m/s²
(d) 10 m/s²

Answer: (a) \( \frac{10}{3} \) m/s²

1. Take g = 10 m/s². If the 20 kg mass moves down by x, the string on both sides of the movable pulley is lengthened by x, so the 10 kg block moves 2x along the table. Hence a₁₀ = 2a, where a is the acceleration of the 20 kg mass.
2. For the 10 kg block: T = 10 × 2a = 20a.
3. For the 20 kg mass (supported by two string segments): 20g - 2T = 20a, so 200 - 40a = 20a.
4. a = 200/60 = 10/3 m/s².
5. Hence, the correct option is (a).

Teacher's Note:
a) A movable pulley doubles the tension on the load and halves the acceleration of the load relative to the free end.
b) Use the string constraint first, then Newton's second law for each mass.

11. The weights w₁ and w₂ are suspended from the ends of a light strig passing over a smooth fixed pulley. If the pulley is pulled up with an acceleration g, the tension in the string will be
(a) \( \frac{w1w2}{w1 + w2} \)
(b) \( \frac{2w1w2}{w1 + w2} \)
(c) \( \frac{4w1w2}{w1 + w2} \)
(d) \( \frac{w1w2}{2(w1 + w2)} \)

Answer: (c) \( \frac{4w1w2}{w1 + w2} \)

1. For a fixed pulley, T = 2w₁w₂/(w₁ + w₂). When the pulley is pulled up with acceleration g, the effective gravity is g + g = 2g.
2. Every weight is effectively doubled, so T = 2 × 2w₁w₂/(w₁ + w₂).
3. T = 4w₁w₂/(w₁ + w₂).
4. Hence, the correct option is (c).

Teacher's Note:
a) This is the same idea as Question 4, with the pulley acceleration equal to g.
b) The tension in a fixed-pulley Atwood machine scales with the effective g.

12. Two blocks each of mass m are connected over two light and frictionless pulleys by a light string as shown in figure. The tension in the string is
(a) 2 mg sin θ
(b) \( \frac{3}{2} \) mg sin θ
(c) \( \frac{2mg sin θ}{3} \)
(d) \( \frac{mg sin θ}{2} \)

Answer: (d) \( \frac{mg sin θ}{2} \)

1. The surfaces are taken as smooth. The block on the incline is pulled down the slope by mg sin θ, and the string pulls the block on the table.
2. Both blocks have the same acceleration a: mg sin θ - T = ma and T = ma.
3. So a = g sin θ/2.
4. T = ma = (mg sin θ)/2.
5. Hence, the correct option is (d).

Teacher's Note:
a) The part of the string that goes down and turns the corner only changes the direction of the tension.
b) Equal masses share the driving force equally.

13. A pendulum is hanging from the ceiling of a railway carriage subtending an angle of 30º with the vertical direction. The acceleration of the carriage is
(a) \( \frac{g}{\sqrt{3}} \)
(b) \( \frac{2g}{\sqrt{3}} \)
(c) g√3
(d) \( \frac{\sqrt{3}g}{2} \)

Answer: (a) \( \frac{g}{\sqrt{3}} \)

1. In the frame of the carriage: T sin 30° = ma and T cos 30° = mg.
2. Dividing, a = g tan 30° = g/√3.
3. Hence, the correct option is (a).

Teacher's Note:
a) The angle with the vertical satisfies tan θ = a/g.
b) Check: for a = g the angle is 45°.

14. A fireman is sliding down with the help of a rope. If the breaking strength of the rope is \( \frac{2mg}{3} \) where m is the mass of the man, the acceleration with which the fireman should slide, so that the rope does not break, is
(a) \( \frac{g}{2} \)
(b) \( \frac{g}{3} \)
(c) \( \frac{g}{4} \)
(d) \( \frac{2g}{3} \)

Answer: (b) \( \frac{g}{3} \)

1. While sliding down with acceleration a, the tension in the rope is T = m(g - a).
2. The rope does not break if T is at most 2mg/3: m(g - a) ≤ 2mg/3.
3. This gives a ≥ g/3.
4. So the minimum acceleration is g/3 and the correct option is (b).

Teacher's Note:
a) A slower slide (smaller a) gives a greater tension.
b) The fireman must slide down with at least g/3.

15. The time taken by a block to slide down a rough inclined plane at 45º to the horizontal is twice as compared to twice the time required by the block to slide down smooth frictionless plane inclined at 45º to the horizontal. The coefficient of kinetic friction between the block and rough plane is
(a) \( \frac{1}{3} \)
(b) \( \sqrt{\frac{3}{4}} \)
(c) \( \frac{3}{4} \)
(d) \( \sqrt{\frac{3}{2}} \)

Answer: (c) \( \frac{3}{4} \)

1. The block is taken to take twice as long on the rough plane as on the smooth plane (the answer options fit this reading), so t_rough = 2t_smooth.
2. For the same distance s = (1/2)at², the acceleration is inversely proportional to t²: a_rough = a_smooth/4.
3. g(sin 45° - μ cos 45°) = (1/4)g sin 45°, so 1 - μ = 1/4.
4. μ = 3/4.
5. Hence, the correct option is (c).

Teacher's Note:
a) At 45°, sin θ = cos θ, so the working simplifies to 1 - μ.
b) The distance is the same in both cases, so compare the accelerations.

16. The force required to just move a body up the inclined plane is twice the force required to just prevent the body from sliding down the plane. If μ is the coefficient of friction, the inclination of the plane to the horizontal is
(a) tan⁻¹ (\( \frac{μ}{2} \))
(b) tan⁻¹ (μ)
(c) tan⁻¹ (2μ)
(d) tan⁻¹ (3μ)

Answer: (d) tan⁻¹ (3μ)

1. Force (along the plane) to just move the body up: F₁ = mg(sin θ + μ cos θ).
2. Force to just prevent it from sliding down: F₂ = mg(sin θ - μ cos θ).
3. F₁ = 2F₂ gives sin θ + μ cos θ = 2 sin θ - 2μ cos θ, so 3μ cos θ = sin θ.
4. tan θ = 3μ, so θ = tan⁻¹(3μ).
5. Hence, the correct option is (d).

Teacher's Note:
a) Friction acts down the slope when the body is about to move up, and up the slope when it is about to slide down.
b) Divide by cos θ to get tan θ.

17. A body of mass m is placed on a horizontal rough surface and is pushed by a force F subtending an angle θ to the horizontal. If the angle of friction is φ, the magnitude of force required to move the body is
(a) \( \frac{mgsinφ}{tan(θ+φ)} \)
(b) \( \frac{mgcosφ}{cos(θ+φ)} \)
(c) \( \frac{mgsinφ}{cos(θ+φ)} \)
(d) \( \frac{mg tan φ}{sin(θ+φ)} \)

Answer: (c) \( \frac{mgsinφ}{cos(θ+φ)} \)

1. The force F pushes the body at an angle θ below the horizontal. The normal reaction N = mg + F sin θ.
2. The body just moves when F cos θ = μN = μ(mg + F sin θ), where μ = tan φ.
3. F(cos θ - μ sin θ) = μmg, so F = μmg/(cos θ - μ sin θ).
4. Putting μ = tan φ: F = mg sin φ/(cos φ cos θ - sin φ sin θ) = mg sin φ/cos(θ + φ).
5. Hence, the correct option is (c).

Teacher's Note:
a) The angle of friction φ is defined by tan φ = μ.
b) The vertical part of the push increases the normal reaction and hence the friction.

18. Two blocks A and B of respective masses 6 kg and 4 kg are connected with a spring passing over a light frictionless pulley as shown in the figure. The coefficient of friction between the block A and horizontal surface is 0.4. Then the minimum mass of block C which should be placed over A to prevent it from moving is
(a) 2 kg
(b) 4 kg
(c) 5 kg
(d) 10 kg

Answer: (b) 4 kg

1. The block B hangs from a string, so the pull on A is the weight of B: 4g.
2. To prevent A from moving, the limiting friction must be at least 4g: μ(m_A + m_C)g ≥ 4g.
3. 0.4(6 + m_C) ≥ 4, so 6 + m_C ≥ 10.
4. m_C ≥ 4 kg.
5. Hence, the correct option is (b).

Teacher's Note:
a) The added mass increases the normal reaction and hence the maximum friction.
b) The word "spring" in the question stands for the string joining A and B.

19. A trolley of mass M is moving on smooth horizontal rails as shown in the figure. It has a rectangular block of mass M' placed in contact with its vertical face as shown. If the coefficient of friction between the block and trolley is μ, then the minimum acceleration a of the trolley, so that the block does not fall, is
(a) μ/g
(b) μg
(c) \( \frac{μ Mg}{M'} \)
(d) \( \frac{g}{μ} \)

Answer: (d) \( \frac{g}{μ} \)

1. In the frame of the trolley, the block is pressed against the vertical face with the normal force N = M'a (the pseudo force).
2. The friction force must balance the weight of the block: μN ≥ M'g.
3. μM'a ≥ M'g, so a ≥ g/μ.
4. The minimum acceleration is g/μ and the correct option is (d).

Teacher's Note:
a) The mass of the block cancels from both sides.
b) A smaller μ needs a larger acceleration to hold the block.

20. A ball of mass 500 gm moving with a speed of 2m/s hits a wall at right angle. It bounces back with the same speed. If the ball had been in contact for one milli-second, the average force exerted by the ball on the wall is
(a) 500 N
(b) 1000 N
(c) 2000 N
(d) 3000 N

Answer: (c) 2000 N

1. Mass m = 0.5 kg. Initial momentum = 0.5 × 2 = 1 kg m/s towards the wall, and final momentum = 1 kg m/s away from it.
2. Change in momentum = 1 - (-1) = 2 kg m/s.
3. Average force = change in momentum/time = 2/(10⁻³) = 2000 N.
4. Hence, the correct option is (c).

Teacher's Note:
a) The change in momentum is twice the initial momentum, because the direction reverses.
b) Convert grams to kilograms and milliseconds to seconds.

21. A force of 3t² Newton acts on a body of mass 10 kg in the forward direction where t is time and a constant frictional force of 32 Newton is present. If the initial speed of the body is 10 m/s, the velocity of the body after 5 second is
(a) 2.5 m/s
(b) 3.5 m/s
(c) 5.0 m/s
(d) 6.5 m/s

Answer: (d) 6.5 m/s

1. The net force on the body = 3t² - 32 N, and F = m dv/dt.
2. Change in velocity = (1/10)∫(3t² - 32) dt from 0 to 5 = (1/10)[t³ - 32t] = (1/10)(125 - 160) = -3.5 m/s.
3. The velocity never falls to zero in between (its least value is about 3 m/s at t = 3.27 s), so friction keeps acting against the motion throughout.
4. v = 10 - 3.5 = 6.5 m/s.
5. Hence, the correct option is (d).

Teacher's Note:
a) Check that the body does not stop in between, otherwise the friction would change.
b) Use impulse = change in momentum with a variable force.

22. A uniform rope of length L resting on a frictionless horizontal surface is pulled at one end by a force F. The tension of the rope at a distance λ from the end where the force is applied is
(a) \( \frac{FL}{L - l} \)
(b) \( \frac{FL}{L + l} \)
(c) \( \frac{F(L - l)}{L} \)
(d) \( \frac{F(L + l)}{L} \)

Answer: (c) \( \frac{F(L - l)}{L} \)

1. The whole rope has acceleration a = F/M, where M is the mass of the rope.
2. The tension at a distance l from the pulled end must accelerate the remaining part of the rope, of length (L - l), whose mass is M(L - l)/L.
3. T = M(L - l)/L × F/M = F(L - l)/L.
4. Hence, the correct option is (c).

Teacher's Note:
a) The tension is greatest at the pulled end and falls to zero at the far end.
b) The symbol λ in the question and the symbol l in the options stand for the same distance.

23. A box of mass M is gently placed over a conveyer belt moving linearly with speed v. The coefficient of friction between the box and conveyer belt is μ. The box will start sliding on the belt after travelling a distance of
(a) \( \frac{v²}{μg} \)
(b) \( \frac{v²}{2μg} \)
(c) \( \frac{v}{μg} \)
(d) \( \frac{v}{2μg} \)

Answer: (b) \( \frac{v²}{2μg} \)

1. The box is placed at rest on the moving belt, so kinetic friction acts on it and gives it an acceleration a = μg.
2. The box speeds up from 0 to the belt speed v after a distance s: v² = 2as = 2μgs, so s = v²/(2μg).
3. After that the box moves along with the belt without sliding.
4. Hence, the correct option is (b).

Teacher's Note:
a) The distance asked for is the distance travelled before the box moves together with the belt (the sliding relative to the belt stops).
b) The mass of the box does not appear in the answer.

24. The blocks B and C in the figure have mass m each. The strings AB and BC are light, having tensions T₁ and T₂ respectively. The system is in equilibrium with a constant horizontal force mg acting on C.
(a) tan θ₁ = 1/2
(b) tan θ₂ = 1
(c) T₁ = √5mg
(d) T₂ = √2mg

Answer: (a) tan θ₁ = 1/2 and (b) tan θ₂ = 1 and (c) T₁ = √5mg and (d) T₂ = √2mg

1. For block C: T₂ cos θ₂ = mg (vertical) and T₂ sin θ₂ = mg (horizontal). So tan θ₂ = 1 and T₂ = √2 mg.
2. For block B: T₁ cos θ₁ = mg + T₂ cos θ₂ = 2mg and T₁ sin θ₁ = T₂ sin θ₂ = mg.
3. So tan θ₁ = 1/2 and T₁ = √((2mg)² + (mg)²) = √5 mg.
4. All four statements are correct, so the correct options are (a), (b), (c) and (d).

Teacher's Note:
a) Resolve the forces on each block into vertical and horizontal parts.
b) The vertical pull on B includes the weight of C carried by the string BC.

25. Two masses m₁ and m₂ are connected by a light string passing over a smooth pulley. When the masses start moving, m₁ moves downwards by 1.4 m in 2 seconds, the ratio m₁/m₂ is (g = 9.8 m/s²)
(a) \( \frac{9}{7} \)
(b) \( \frac{11}{9} \)
(c) \( \frac{13}{11} \)
(d) \( \frac{15}{13} \)

Answer: (d) \( \frac{15}{13} \)

1. From s = (1/2)at²: 1.4 = (1/2) × a × 4, so a = 0.7 m/s².
2. For an Atwood machine, a = g(m₁ - m₂)/(m₁ + m₂), so (m₁ - m₂)/(m₁ + m₂) = 0.7/9.8 = 1/14.
3. Applying componendo and dividendo: m₁/m₂ = (14 + 1)/(14 - 1) = 15/13.
4. Hence, the correct option is (d).

Teacher's Note:
a) Componendo and dividendo quickly gives the ratio of the masses.
b) Check: 15 - 13 = 2 and 15 + 13 = 28, so the ratio is 2/28 = 1/14.

26. A particle is projected up a 45º rough incline with a velocity v'. The coefficient of friction is 0.5. The speed with which it returns back to the starting point is v. Then v' /v is
(a) 1.3
(b) 1/√3
(c) 1/2
(d) 1/√2

Answer: (b) 1/√3

1. Going up, the retardation is a₁ = g(sin 45° + μ cos 45°) = 1.5g sin 45°. Coming down, a₂ = g(sin 45° - μ cos 45°) = 0.5g sin 45°.
2. The distance travelled up is s = v'²/(2a₁), and the speed on return is v² = 2a₂s = v'² a₂/a₁ = v'²/3.
3. So v/v' = 1/√3 (and v'/v = √3).
4. The ratio 1/√3 given in option (b) is the ratio of the returning speed to the projection speed, so the correct option is (b).

Teacher's Note:
a) The ratio printed in the question as v'/v is the other way round; the option 1/√3 corresponds to v/v'.
b) The block returns with a lower speed because of the work done against friction in both directions.

27. A uniform chain of length L lies on a table. If the coefficient of friction is μ, then the maximum length of the chain which can hang from the edge of the table without the chain sliding down is
(a) \( \frac{L}{μ} \)
(b) \( \frac{L}{μ - 1} \)
(c) \( \frac{μL}{μ + 1} \)
(d) \( \frac{μL}{μ - 1} \)

Answer: (c) \( \frac{μL}{μ + 1} \)

1. Let the length l hang from the edge. The weight of the hanging part = (l/L)Mg and the maximum friction on the part on the table = μ((L - l)/L)Mg.
2. For equilibrium: l = μ(L - l).
3. l(1 + μ) = μL, so l = μL/(μ + 1).
4. Hence, the correct option is (c).

Teacher's Note:
a) The hanging part pulls the chain down; friction on the part lying on the table resists it.
b) Check: for μ = 1, half of the chain can hang.

28. A 1000 kg lift is supported by a cable that can support 2000 kg. The shortest distance in which the lift can be stopped when it is descending with a speed of 2.5 m/s is (g = 10 m/s²)
(a) \( \frac{5}{16} \) m
(b) \( \frac{5}{32} \) m
(c) 1m
(d) 2m

Answer: (a) \( \frac{5}{16} \) m

1. The maximum tension in the cable = 2000g = 20000 N.
2. When the lift is being stopped, the net upward force is T - mg = 20000 - 10000 = 10000 N, so a = 10000/1000 = 10 m/s² (retardation).
3. s = v²/(2a) = (2.5)²/(2 × 10) = 6.25/20 = 5/16 m.
4. Hence, the correct option is (a).

Teacher's Note:
a) The cable can be loaded to its limit, which gives the largest possible retardation and so the shortest stopping distance.
b) Subtract the weight of the lift to get the net force.

29. In the given diagram, with what force must the man pull the rope to hold the plank in position? Weight of the man is 60 kgf. Neglect the weights of plank, rope and pulley.
(a) 15 kgf
(b) 30 kgf
(c) 60 kgf
(d) 120 kgf

Answer: (a) 15 kgf

1. Let T be the tension in the rope held by the man. The same rope passes round the movable pulley, so the pulley is pulled up by 2T.
2. The rope over the fixed pulley carries the tension 2T on both sides, so it pulls up the plank with 2T on the left.
3. Taking man and plank as one system, the upward forces are 2T (left rope), T (the end of the rope fixed to the plank) and T (the rope held by the man).
4. Total upward force 4T = weight of man = 60 kgf, so T = 15 kgf.
5. Hence, the correct option is (a).

Teacher's Note:
a) Count the rope segments that support the man and plank together.
b) A movable pulley halves the force needed.

30. A block of mass 50 kg can slide on a rough horizontal surface. The coefficient of friction between the block and the surface is 0.6. The least force of pull, acting 30º to the upward drawn vertical, which causes the block to just slide is (g = 9.8 m/s²)
(a) 288.3 N
(b) 215.2N
(c) 219.6N
(d) 294.2 N

Answer: (a) 288.3 N

1. The pull F makes 30° with the upward vertical, so its horizontal component is F sin 30° and its vertical component is F cos 30°.
2. Normal reaction N = mg - F cos 30°.
3. For the block to just slide: F sin 30° = μ(mg - F cos 30°), so F(0.5 + 0.6 × 0.866) = 0.6 × 50 × 9.8.
4. F × 1.0196 = 294, so F = 288.3 N.
5. Hence, the correct option is (a).

Teacher's Note:
a) A pull with an upward component reduces the normal reaction, and so the friction.
b) Be careful to measure the angle from the vertical, not from the horizontal.

(a) 2 mg sin θ (b) 3/2 mg sin θ m θ m (c) 2mgsin θ/3  (d) mgsinθ/2