JEE Physics Laws Of Motion MCQs Set 10

Answer: (d) none of the above

Explanation:
1. The coefficient of friction depends on the nature (material and roughness) of the two surfaces.
2. It does not depend on the normal reaction, the shape or the area of contact.

Teacher's Note:
1. Friction force depends on N, but the coefficient does not.
2. The nature of the surfaces fixes the coefficient.

Answer: (b) upwards

Explanation:
1. The book tends to slide down because of its weight.
2. Friction opposes the tendency of motion, so it acts upwards.

Teacher's Note:
1. Friction always opposes the relative motion or its tendency.
2. The upward friction balances the weight.

Answer: (c) upward and tilted opposite to the direction of motion of the train

Explanation:
1. The box retards along with the train, so the net horizontal force on it is opposite to the motion, and this force is the friction from the floor.
2. The normal reaction acts upward.
3. The resultant reaction is upward and tilted opposite to the train's motion.

Teacher's Note:
1. The total reaction is the vector sum of normal reaction and friction.
2. Friction points in the direction of the acceleration of the box.

Answer: (b) 8.0 N

Explanation:
1. Limiting friction \( = 0.4 \times 4 \times 10 = 16 \) N.
2. The applied force 8 N is less than 16 N, so the block does not move.
3. Static friction equals the applied force, 8 N.

Teacher's Note:
1. Compare the applied force with the limiting friction.
2. If it is smaller, friction equals the applied force.

Answer: (a) 8.7 N

Explanation:
1. Limiting friction \( = 0.4 \times 3 \times 10 = 12 \) N.
2. The applied force 8.7 N is less than 12 N, so the block stays at rest.
3. Friction equals the applied force, 8.7 N.

Teacher's Note:
1. A block at rest has static friction equal to the applied force.
2. Check the limiting value first.

Answer: (b) 0.5 N

Explanation:
1. \( mg - f = ma \).
2. \( f = m(g - a) = 0.5 \times (10 - 9) = 0.5 \) N.

Teacher's Note:
1. The air resistance reduces the acceleration below g.
2. Net downward force is weight minus resistance.

Answer: (a) 1000 m

Explanation:
1. Maximum retardation \( a = \mu_k g = 0.5 \times 10 = 5 \) m/s².
2. \( s = \dfrac{v^2}{2a} = \dfrac{100^2}{2 \times 5} = 1000 \) m.

Teacher's Note:
1. Friction gives the retardation \( \mu g \).
2. Then use \( v^2 = 2as \).

Answer: (c) 5 N

Explanation:
1. The maximum acceleration friction can give is \( \mu g = 2 \) m/s², which is more than 0.5 m/s².
2. So the block does not slip and moves with the trolley.
3. Friction \( = ma = 10 \times 0.5 = 5 \) N.

Teacher's Note:
1. First check whether slipping occurs.
2. If not, friction equals mass times the trolley's acceleration.

Answer: (d) 0.375

Explanation:
1. Limiting static friction \( = 75 \) N, and \( N = mg = 200 \) N.
2. \( \mu_s = 75/200 = 0.375 \).
3. (The 60 N gives the kinetic coefficient 0.3.)

Teacher's Note:
1. The force that just starts motion equals limiting static friction.
2. The force for uniform motion gives the kinetic coefficient.

Answer: (d) 0.8

Explanation:
1. \( F - f = 10 \times 5 \) and \( 2F - f = 10 \times 18 \).
2. Subtracting: \( F = 180 - 50 = 130 \) N.
3. Then \( f = 130 - 50 = 80 \) N and \( \mu = f/mg = 80/100 = 0.8 \).

Teacher's Note:
1. Write the second law for both cases and subtract.
2. Friction stays the same in both cases.

Answer: (d) the system cannot be in equilibrium

Explanation:
1. The only upward force that could support the blocks is friction.
2. The wall is smooth, so it gives no friction.
3. Friction between A and B is an internal force, so the weight of the system cannot be balanced, and the system cannot be in equilibrium.

Teacher's Note:
1. Only external forces can balance the weight of the system.
2. Internal friction between A and B cancels out.

Answer: (c) 20 N downwards

Explanation:
1. Normal reaction \( N = 100\cos30° = 50\sqrt{3} \) N, so the limiting friction is \( \mu N = 50 \) N.
2. Vertical upward component of F \( = 100\sin30° = 50 \) N, and the weight is 30 N.
3. The block is pushed up with a net 20 N, so friction of 20 N acts downwards (less than 50 N, so the block stays at rest).

Teacher's Note:
1. Find the direction of the tendency to slide before choosing the direction of friction.
2. Check that the needed friction is within the limiting value.

Answer: (b) \( \dfrac{mg}{\sin\theta + \mu\cos\theta} \)

Explanation:
1. Normal reaction \( N = F\cos\theta \).
2. For the minimum F, friction \( \mu N \) acts upward along with \( F\sin\theta \): \( F\sin\theta + \mu F\cos\theta = mg \).
3. \( F = \dfrac{mg}{\sin\theta + \mu\cos\theta} \).

Teacher's Note:
1. For the minimum force friction is at its limiting value acting upward.
2. Resolve F along the wall and perpendicular to it.

Answer: (b) \( \dfrac{mg}{\sin\theta} \)

Explanation:
1. With zero friction, the vertical component of F alone balances the weight.
2. \( F\sin\theta = mg \), so \( F = mg/\sin\theta \).

Teacher's Note:
1. Zero friction means the weight is balanced by the vertical component of F alone.
2. The wall's normal reaction balances the horizontal component.

Answer: (c) 18.0 N

Explanation:
1. Normal reaction \( N = 200 + 30\sin53° = 200 + 24 = 224 \) N, so the limiting friction is \( 0.2 \times 224 = 44.8 \) N.
2. The horizontal component of F is \( 30\cos53° = 18 \) N, which is less than 44.8 N.
3. So the block does not move and static friction is 18 N.

Teacher's Note:
1. Compare the horizontal pull with the limiting friction.
2. A downward-slanted force increases the normal reaction.

Answer: (d) 1.2 kg

Explanation:
1. Limiting friction on A \( = 0.3 \times 4 \times 10 = 12 \) N.
2. The hanging block just fails to move the system when \( m_B g = 12 \).
3. So \( m_B = 1.2 \) kg.

Teacher's Note:
1. The weight of B must not exceed the limiting friction on A.
2. Find the limiting friction first.

Answer: (a) 1/4

Explanation:
1. Let the rope have length L and mass M. Hanging weight \( = (M/5)g \).
2. Friction \( = \mu \times (4M/5)g \) from the part on the table.
3. At the point of sliding, \( \tfrac15 = \mu\cdot\tfrac45 \), so \( \mu = 1/4 \).

Teacher's Note:
1. The hanging part provides the pulling force.
2. The part lying on the table gives the normal reaction.

Answer: (a) 2 N

Explanation:
1. The maximum acceleration friction can give \( = \mu g = 2 \) m/s², which is more than 1 m/s².
2. So the block does not slip and moves with the plank.
3. Friction \( = ma = 2 \times 1 = 2 \) N.

Teacher's Note:
1. Check that the block does not slip before using friction equal to ma.
2. If it slips, use kinetic friction.

Answer: (b) m(g + a)

Explanation:
1. The friction force of the rope on the monkey acts upward.
2. Newton's second law: \( f - mg = ma \).
3. So \( f = m(g + a) \).

Teacher's Note:
1. The friction is static, so it is found from Newton's law and not from μ times a normal force.
2. Use the net upward force.

Answer: (d) \( \tan\theta \)

Explanation:
1. Constant speed means zero net force along the plane.
2. \( mg\sin\theta = \mu_k mg\cos\theta \).
3. So \( \mu_k = \tan\theta \).

Teacher's Note:
1. Constant speed means balanced forces.
2. The coefficient of friction is a pure number.

Answer: (d) \( \dfrac{2v_0^2}{\mu g} \)

Explanation:
1. Maximum retardation \( = \mu g \).
2. \( s = \dfrac{(2v_0)^2}{2\mu g} = \dfrac{2v_0^2}{\mu g} \).

Teacher's Note:
1. Maximum friction gives the shortest stopping distance.
2. Use \( v^2 = 2as \).

Answer: (c) \( \tan^{-1}(2\mu) \)

Explanation:
1. Force to move up: \( mg(\sin\theta + \mu\cos\theta) \). Force to prevent sliding down: \( mg(\sin\theta - \mu\cos\theta) \).
2. Setting the first equal to three times the second: \( \sin\theta + \mu\cos\theta = 3\sin\theta - 3\mu\cos\theta \).
3. So \( 4\mu\cos\theta = 2\sin\theta \) and \( \tan\theta = 2\mu \).

Teacher's Note:
1. Friction reverses direction between the two cases.
2. Write both force expressions before comparing.

Answer: (d) 5.4

Explanation:
1. \( a = g(\sin45° - \mu\cos45°) = 10 \times 0.707 \times (1 - 0.25) \).
2. \( a \approx 5.3 \) m/s², and the nearest value among the options is 5.4.

Teacher's Note:
1. Acceleration down a rough incline is \( g(\sin\theta - \mu\cos\theta) \).
2. Use \( \sin45° = \cos45° = 0.707 \).

Answer: (a) \( mg\sin\theta + \mu mg\cos\theta \)

Explanation:
1. Moving up the plane, friction acts down the plane.
2. The force must balance \( mg\sin\theta \) and the friction \( \mu mg\cos\theta \).
3. \( F_1 = mg\sin\theta + \mu mg\cos\theta \).

Teacher's Note:
1. Friction opposes the motion, so it points down the plane.
2. The minimum force to just move the block is the sum.