JEE Physics Laws Of Motion MCQs Set 22

Answer: (c) 5 m

Explanation:
1. Retardation = \( g\sin37^\circ+\mu g\cos37^\circ=6+4=10 \) m/s².
2. \( s=\dfrac{v^2}{2a}=\dfrac{100}{20}=5 \) m.

Teacher's Note:
1. Both gravity and friction oppose the upward motion.
2. Friction reverses direction when the block starts to come down.

Answer: (d) \( t_0=\dfrac{\mu g\,m_2(m_1+m_2)}{k\,m_1} \)

Explanation:
1. Together: \( a=\dfrac{kt}{m_1+m_2} \), and the plank's acceleration comes only from friction: \( m_1a\le\mu m_2g \).
2. Slipping starts when \( \dfrac{m_1kt_0}{m_1+m_2}=\mu m_2g \).

Teacher's Note:
1. The plank is accelerated only by friction from the bar.
2. After \( t_0 \) the plank's acceleration stays constant at \( \mu m_2g/m_1 \).

Answer: (b) the static friction of the ground on your feet

Explanation:
1. Your feet do not slip on the ground, so the force along the ground that stops you is friction.
2. Because there is no slipping, it is static friction.

Teacher's Note:
1. Normal reaction is perpendicular to the ground and cannot change horizontal motion.
2. Friction can also be the cause of acceleration, not only of retardation.

Answer: (c) 1 m/s²

Explanation:
1. Force applied = \( \mu_smg=0.5\times60\times9.8=294 \) N.
2. Sliding friction = \( 0.4\times60\times9.8=235 \) N.
3. \( a=\dfrac{294-235.2}{60}=0.98\approx1 \) m/s².

Teacher's Note:
1. Once motion starts, the smaller kinetic friction acts.
2. The net force is the difference between the two friction values.

Answer: (a) 80 N

Explanation:
1. Friction A–B = \( 0.25\times30=7.5 \) N, B–C = \( 0.25\times70=17.5 \) N and C–floor = \( 0.25\times150=37.5 \) N.
2. For B at constant velocity: \( T=7.5+17.5=25 \) N.
3. For C: \( F=T+17.5+37.5=80 \) N.

Teacher's Note:
1. Normal reactions add up as the stack is built down from A.
2. The string tension acts on both B and C.

Answer: (d) 10 m

Explanation:
1. \( N=100 \) N, friction \( =0.5\times100=50 \) N, so deceleration = 5 m/s².
2. \( 0=100-2(5)s \), so s = 10 m.

Teacher's Note:
1. Mass cancels in this kind of problem.
2. Use \( v^2=u^2+2as \).

Answer: (d) 15 kg

Explanation:
1. Tension = weight of B = 5g.
2. Limiting friction must at least equal this: \( 0.2(10+m)g\ge5g \).
3. \( 10+m\ge25 \), so m = 15 kg.

Teacher's Note:
1. The normal reaction on the table includes the extra block.
2. Set friction equal to the pull of the hanging weight.

Answer: (d) 15 N

Explanation:
1. There is no friction between B and C, so A and B accelerate together, independent of C.
2. A is dragged only by friction, so \( a_{max}=\mu g=2.5 \) m/s².
3. \( F=(m_A+m_B)a_{max}=6(2.5)=15 \) N.

Teacher's Note:
1. C's friction with the ground is irrelevant because B slides freely over C.
2. Sliding starts when A's required acceleration exceeds the maximum friction can give.

Answer: (d) \( \dfrac{\mu^2a}{4} \)

Explanation:
1. At equilibrium on a wire inclined at angle \( \theta \), \( \tan\theta\le\mu \).
2. Slope \( \dfrac{dy}{dx}=\dfrac{2x}{a}=\tan\theta=\mu \), so \( x=\dfrac{\mu a}{2} \).
3. \( y=\dfrac{x^2}{a}=\dfrac{\mu^2a}{4} \).

Teacher's Note:
1. The limiting angle is the angle of friction.
2. The slope of the wire increases with x, so equilibrium is possible up to a maximum x.

Answer: (c) 0.05g

Explanation:
1. \( m_2g-T=m_2a \) and \( T-m_1g\sin\alpha-km_1g\cos\alpha=m_1a \).
2. \( a=\dfrac{g(\eta-\sin\alpha-k\cos\alpha)}{\eta+1}=\dfrac{g(0.667-0.5-0.0866)}{1.667} \).
3. \( a\approx0.048g\approx0.05g \).

Teacher's Note:
1. Add the two equations to eliminate the tension.
2. Positive result confirms that m2 moves down.

Answer: (a) 4 m/s

Explanation:
1. When the chain has moved x, the rough part has length \( L-x \), so friction = \( \mu_k\lambda g(L-x) \).
2. Work done = \( FL-\tfrac12\mu_k\lambda gL^2=21-5=16 \) J.
3. \( \tfrac12(\lambda L)v^2=16 \) gives \( v^2=16 \), so v = 4 m/s.

Teacher's Note:
1. Friction decreases linearly as the chain leaves the rough part.
2. Use the work-energy theorem for the whole chain.

Answer: (b) 5

Explanation:
1. Pushing up: \( F_1=mg\sin45^\circ+\mu mg\cos45^\circ \). Preventing sliding: \( F_2=mg\sin45^\circ-\mu mg\cos45^\circ \).
2. \( F_1=3F_2 \) gives \( 1+\mu=3(1-\mu) \), so \( \mu=0.5 \).
3. k = 10μ = 5.

Teacher's Note:
1. At 45° the sine and cosine are equal and cancel from both sides.
2. Friction acts down the plane for the upward push and up the plane to prevent sliding.

Answer: (c) 5 m/s²

Explanation:
1. In the plane of the wall the forces are the weight 20 N (down) and 15 N (horizontal); resultant = 25 N.
2. Limiting friction = \( 0.5\times30=15 \) N, which opposes the resultant.
3. \( a=\dfrac{25-15}{2}=5 \) m/s².

Teacher's Note:
1. The 30 N force pressing on the wall fixes the normal reaction and so the friction.
2. Add the two forces in the wall plane as vectors.

Answer: (b) 5 N

Explanation:
1. Limiting friction = \( 0.6\times1\times10=6 \) N.
2. The block needs \( ma=5 \) N to accelerate with the truck, which is below 6 N, so it does not slip.
3. The friction is 5 N.

Teacher's Note:
1. Static friction adjusts itself up to the limiting value.
2. Do not use the limiting value unless slipping is just about to start.

Answer: (b) \( \tan\beta=k \)

Explanation:
1. Along the incline: \( T\cos\beta=mg\sin\alpha+k(mg\cos\alpha-T\sin\beta) \).
2. \( T=\dfrac{mg(\sin\alpha+k\cos\alpha)}{\cos\beta+k\sin\beta} \), least when the denominator is largest.
3. The denominator is maximum when \( \tan\beta=k \).

Teacher's Note:
1. The denominator \( \cos\beta+k\sin\beta \) has maximum \( \sqrt{1+k^2} \).
2. The best angle is the angle of friction.

Answer: (d) \( \dfrac{m+n}{n}\tan\theta \)

Explanation:
1. Speed at the end of the smooth part: \( v^2=2g\sin\theta\,m \).
2. On the rough part: \( 0=v^2+2g(\sin\theta-\mu\cos\theta)n \).
3. \( 2gm\sin\theta+2gn\sin\theta=2g\mu n\cos\theta \), so \( \mu=\dfrac{m+n}{n}\tan\theta \).

Teacher's Note:
1. The work done by gravity over the whole length equals the work against friction over the rough length.
2. Check: if m = 0, then \( \mu=\tan\theta \).

Answer: (b) 2

Explanation:
1. Tension = 100g = 1000 N (g = 10).
2. Friction must be 1000 N: \( \mu(50\times10)=1000 \), so \( \mu=2 \).

Teacher's Note:
1. Coefficients of friction can be greater than 1.
2. The normal reaction on the table block is only 500 N.

Answer: (b) \( \dfrac{\tan\theta}{\mu_0} \)

Explanation:
1. Acceleration \( a=g\sin\theta-\mu_0xg\cos\theta \).
2. Speed is maximum when \( a=0 \): \( x=\dfrac{\tan\theta}{\mu_0} \).

Teacher's Note:
1. Maximum speed occurs when acceleration becomes zero.
2. Friction grows linearly with distance until it equals the driving component.

Answer: (a) Only (1) and (2)

Explanation:
1. Weight = 1 N. Component of weight down the slope = \( \sin\theta \); component of the 1 N force up the slope = \( \cos\theta \).
2. At 45° they are equal and friction is zero. For \( \theta>45^\circ \) the block tends to slide down, so friction acts up the slope, towards Q.
3. For \( \theta<45^\circ \) it tends to move up the slope, so friction would act towards P, not Q.

Teacher's Note:
1. Resolve both forces along the plane first, then give friction the opposite direction.
2. The plane is at the angle where both components are equal at 45°.

Answer: (a) 20 N

Explanation:
1. Horizontal: \( F\cos60^\circ\le\mu N \). Vertical: \( N=mg+F\sin60^\circ=10\sqrt3+\dfrac{\sqrt3}{2}F \).
2. \( \dfrac F2=\dfrac{1}{2\sqrt3}\left(10\sqrt3+\dfrac{\sqrt3F}2\right)=5+\dfrac F4 \).
3. \( \dfrac F4=5 \), so F = 20 N.

Teacher's Note:
1. Pushing downward increases the normal reaction and so the friction.
2. The block just stays in place at the limiting value.

Answer: (a) 4 m/s²

Explanation:
1. No-slip would need \( a=120/25=4.8 \) m/s² and friction \( 120-48=72 \) N, above the limit 60 N, so slipping occurs.
2. Friction on the 15 kg block is 60 N (the string forces on its pulleys cancel horizontally).
3. \( a=60/15=4 \) m/s².

Teacher's Note:
1. Once slipping starts, kinetic friction (taken as 60 N) acts on both blocks.
2. The 10 kg block accelerates at 6 m/s² in this case.

Answer: (c) 2.1 m/s²

Explanation:
1. Limiting static friction = \( 0.5\times100=50 \) N, and 51 N exceeds it, so the block moves.
2. Kinetic friction = \( 0.3\times100=30 \) N.
3. \( a=\dfrac{51-30}{10}=2.1 \) m/s².

Teacher's Note:
1. Once motion starts, use the kinetic coefficient.
2. If the force were 50 N or less, the acceleration would be zero.