JEE Physics Laws Of Motion MCQs Set 13

Answer: (a) 0.11

Explanation:
1. From \( s=\frac12at^2 \): \( 8=\frac12a(2)^2\Rightarrow a=4\ \text{m/s}^2 \).
2. Along the incline \( a=g\sin\theta-\mu g\cos\theta \), so \( 4=5-\mu(10)(0.866) \).
3. \( \mu=\frac{1}{8.66}\approx0.11 \).

Teacher's Note:
1. Resolve weight into components along and perpendicular to the plane first.
2. Kinetic friction equals mu times the normal reaction, which is mg cos(theta) on a plane.

Answer: (c) 13 N, 0 N

Explanation:
1. Component of weight along the plane: \( mg\sin30^\circ=10\ \text{N} \). Limiting friction: \( \mu mg\cos30^\circ=0.2\times17.3=3.5\ \text{N} \).
2. To move up the plane the force must overcome both: \( F=10+3.5\approx13\ \text{N} \).
3. Since 10 N is already greater than the maximum friction 3.5 N, the block slides down by itself, so no force is needed for (b): 0 N.

Teacher's Note:
1. Always compare the weight component along the plane with the limiting friction before deciding the direction of motion.
2. If the weight component exceeds limiting friction, the block moves down without any push.

Answer: (d) 2.4 m/s²

Explanation:
1. If free, the 2.0 kg block would have \( a_1=g(\sin30^\circ-0.3\cos30^\circ)=10(0.5-0.26)=2.4\ \text{m/s}^2 \).
2. If free, the 4.0 kg block would have \( a_2=10(0.5-0.2\times0.866)=3.27\ \text{m/s}^2 \).
3. The lower block tends to accelerate faster and moves away, so the blocks lose contact and there is no force between them.
4. Hence the 2.0 kg block accelerates at 2.4 m/s².

Teacher's Note:
1. Check whether the rear block would push the front block: it does so only if the rear block's free acceleration is larger.
2. Here the front block is faster, so the blocks separate.

Answer: (a) \( \frac13\tan\theta \)

Explanation:
1. Just preventing downward motion: friction acts up the plane, \( F+\mu mg\cos\theta=mg\sin\theta \).
2. Just sliding up: friction acts down the plane, \( 2F=mg\sin\theta+\mu mg\cos\theta \).
3. Eliminate F: \( 2mg\sin\theta-2\mu mg\cos\theta=mg\sin\theta+\mu mg\cos\theta \), so \( \sin\theta=3\mu\cos\theta \).
4. \( \mu=\frac13\tan\theta \).

Teacher's Note:
1. Friction always opposes the tendency of motion, so its direction flips between the two cases.
2. Subtracting the equations eliminates the unknown force F.

Answer: (c) 16 m

Explanation:
1. Maximum acceleration friction can give the box: \( \mu g=1.5\ \text{m/s}^2 \), which is less than 2 m/s², so the box slips backwards relative to the truck.
2. Relative acceleration of the truck with respect to the box: \( 2-1.5=0.5\ \text{m/s}^2 \).
3. The box must slip 4 m relative to the truck: \( 4=\frac12(0.5)t^2\Rightarrow t=4\ \text{s} \).
4. Distance moved by the truck (the starting point of the truck): \( \frac12(2)(4)^2=16\ \text{m} \).

Teacher's Note:
1. Use relative acceleration when two bodies move with different accelerations along the same line.
2. The question asks about the truck's distance from the starting point, so use the truck's acceleration for the final step.

Answer: (c) \( \frac{\mu l}{\mu+1} \)

Explanation:
1. Let the hanging length be x. Its weight pulls the rope: \( \lambda xg \) where \( \lambda \) is mass per unit length.
2. The friction on the part on the table limits this: \( \mu\lambda(l-x)g \).
3. At the limit \( x=\mu(l-x)\Rightarrow x=\frac{\mu l}{\mu+1} \).

Teacher's Note:
1. Treat mass per unit length as constant so it cancels.
2. The normal reaction on the table part equals its own weight only.

Answer: (c) \( \frac{g}{\sqrt2}\left(1+\frac12\right) \)

Explanation:
1. While moving up, gravity's component and friction both act down the plane.
2. \( a=g\sin45^\circ+\mu g\cos45^\circ=\frac{g}{\sqrt2}(1+\mu) \).
3. With \( \mu=\frac12 \) the retardation is \( \frac{g}{\sqrt2}\left(1+\frac12\right) \).

Teacher's Note:
1. Going up, friction and gravity add; going down, they oppose each other.
2. At 45 degrees, sin and cos are equal, which simplifies the algebra.

Answer: (a) 1.764 N

Explanation:
1. Maximum acceleration that friction can give B: \( a_{max}=\mu g=0.15\times9.8=1.47\ \text{m/s}^2 \).
2. If both move together with this acceleration: \( F=(1+0.2)(1.47)=1.764\ \text{N} \).

Teacher's Note:
1. Check the upper body: friction is the only horizontal force on it.
2. Beyond the limiting acceleration, the two bodies no longer move together.

Answer: (a) 20%

Explanation:
1. Let the hanging fraction be f. Balance: \( f=\mu(1-f) \).
2. \( f=\frac{\mu}{1+\mu}=\frac{0.25}{1.25}=0.2 \), that is 20%.

Teacher's Note:
1. The same result as for a rope: the hanging fraction is mu divided by (1 + mu).
2. Limiting equilibrium means the hanging weight equals the maximum friction.

Answer: (d) 2 s

Explanation:
1. At the limiting angle \( \tan\theta=\mu_s=0.75 \), so \( \sin\theta=0.6,\ \cos\theta=0.8 \).
2. Sliding acceleration: \( a=g(\sin\theta-\mu_k\cos\theta)=10(0.6-0.4)=2\ \text{m/s}^2 \).
3. \( 4=\frac12(2)t^2\Rightarrow t=2\ \text{s} \).

Teacher's Note:
1. The angle of repose satisfies tan(theta) = mu_s.
2. After sliding starts, use the kinetic coefficient.

Answer: (a) 206 N

Explanation:
1. Resolve along the plane: \( F\cos30^\circ=mg\sin30^\circ+f \).
2. \( F\cos30^\circ=20\times9.8\times0.5+80=178\ \text{N} \).
3. \( F=\frac{178}{0.866}\approx206\ \text{N} \).

Teacher's Note:
1. A horizontal force has a component along the plane equal to F cos(theta), not F.
2. Zero acceleration means the forces along the plane balance.

Answer: (a) \( \sqrt{\frac{gR}{\mu_s}} \)

Explanation:
1. The wall's normal reaction supplies the centripetal force: \( N=\frac{mv^2}{R} \).
2. Friction balances weight: \( \mu_sN\ge mg \).
3. \( \mu_s\frac{mv^2}{R}=mg\Rightarrow v_{min}=\sqrt{\frac{gR}{\mu_s}} \).

Teacher's Note:
1. Friction acts vertically upward here, and the normal force is horizontal.
2. A larger coefficient of friction lowers the minimum speed.

Answer: (a) \( \left(\frac{2-\sqrt3}{4}\right)g \)

Explanation:
1. \( a=g(\sin30^\circ-\mu\cos30^\circ)=g\left(\frac12-\frac{\sqrt3}{4}\right) \).
2. \( a=\left(\frac{2-\sqrt3}{4}\right)g \).

Teacher's Note:
1. Sliding without rolling means we treat the ball as a block.
2. The mass cancels out.

Answer: (a) \( \tan\theta\left(1-\frac1{n^2}\right) \)

Explanation:
1. For the same distance from rest, \( s=\frac12at^2 \) gives \( a_{rough}=\frac{a_{smooth}}{n^2} \).
2. \( g\sin\theta-\mu g\cos\theta=\frac{g\sin\theta}{n^2} \).
3. \( \mu=\tan\theta\left(1-\frac1{n^2}\right) \).

Teacher's Note:
1. Same distance and starting from rest means acceleration is inversely proportional to the square of time.
2. Divide by g cos(theta) at the end to isolate mu.

Answer: (a) \( 1-\frac1{n^2} \)

Explanation:
1. Acceleration is inversely proportional to the square of the time, so \( a_{rough}=\frac{a_{smooth}}{n^2} \).
2. \( g\sin45^\circ(1-\mu)=\frac{g\sin45^\circ}{n^2} \) because sin45° = cos45°.
3. \( \mu=1-\frac1{n^2} \).

Teacher's Note:
1. At 45 degrees the factor sin(theta) = cos(theta) cancels completely.
2. This is the special case of the previous kind of problem with tan(45) = 1.

Answer: (c) 75 m/s

Explanation:
1. At the start (v = 0): \( a_0=\frac{100-25}{15}=5\ \text{m/s}^2 \).
2. Half of that is 2.5 m/s², so net force \( =15\times2.5=37.5\ \text{N} \).
3. \( 100-25-0.5v=37.5\Rightarrow v=75\ \text{m/s} \).

Teacher's Note:
1. Write the net force as applied force minus the velocity-dependent resistance.
2. Compute the initial acceleration first, because the target is relative to it.

Answer: (b) 0.6 m/s²

Explanation:
1. Treat the two blocks as one system of mass 5 kg.
2. Net force: \( 5-2=3\ \text{N} \).
3. \( a=\frac{3}{5}=0.6\ \text{m/s}^2 \).

Teacher's Note:
1. Use a single system when both bodies move with the same acceleration.
2. Internal contact forces cancel and do not appear in the system equation.

Answer: (b) M/3

Explanation:
1. Constant speed means zero acceleration for both blocks.
2. For A: \( Mg\sin30^\circ-\mu Mg\cos30^\circ=T \) and for B: \( T=m_Bg \).
3. \( m_B=M\left(\frac12-\frac1{3\sqrt3}\cdot\frac{\sqrt3}{2}\right)=M\left(\frac12-\frac16\right)=\frac M3 \).

Teacher's Note:
1. At constant velocity the net force on each block is zero.
2. Friction acts up the plane because A moves down.

Answer: (a) 1.3 m/s²

Explanation:
1. Taken together, driving force \( =(4+2)(9.8)(0.6)=35.28\ \text{N} \).
2. Total friction \( =0.75(4)(9.8)(0.8)+0.25(2)(9.8)(0.8)=23.52+3.92=27.44\ \text{N} \).
3. \( a=\frac{35.28-27.44}{6}=1.3\ \text{m/s}^2 \). (Alone, M1 would not move at all, while M2 would accelerate, so the string stays taut.)

Teacher's Note:
1. The string stays tight because the lower block would run ahead on its own.
2. Treat the pair as a single system to find the common acceleration.

Answer: (a) Zero

Explanation:
1. In a freely falling lift the effective gravity is \( g_{eff}=g-g=0 \).
2. Then the block has no apparent weight, so there is no normal reaction and no friction, and nothing makes it slide.
3. The acceleration relative to the incline is zero.

Teacher's Note:
1. Free fall means weightlessness inside the lift.
2. Without normal force there can be no friction, and without weight there is no driving component.

Answer: (a) \( \frac FM(\cos\phi+\mu\sin\phi)-\mu g \)

Explanation:
1. Normal reaction: \( N=Mg-F\sin\phi \).
2. Horizontal: \( F\cos\phi-\mu(Mg-F\sin\phi)=Ma \).
3. \( a=\frac FM(\cos\phi+\mu\sin\phi)-\mu g \).

Teacher's Note:
1. The upward component of the pull reduces the normal reaction.
2. Less normal reaction means less friction.

Answer: (a) \( \frac{mMg}{nm+M} \)

Explanation:
1. Whole system accelerates with \( a=\frac{Mg}{nm+M} \).
2. The string on the n-th block pulls only that block of mass m: \( T=ma=\frac{mMg}{nm+M} \).

Teacher's Note:
1. Only the last block lies on the free end, so the tension equals its own mass times the common acceleration.
2. Find the system acceleration first.

Answer: (d) \( \left[\frac{2L}{(g+a_0)\sin\theta\cos\theta}\right]^{1/2} \)

Explanation:
1. In the lift frame, effective gravity is \( g+a_0 \), so the acceleration along the incline is \( (g+a_0)\sin\theta \).
2. The length of the incline is \( \frac{L}{\cos\theta} \).
3. \( \frac{L}{\cos\theta}=\frac12(g+a_0)\sin\theta\,t^2\Rightarrow t=\left[\frac{2L}{(g+a_0)\sin\theta\cos\theta}\right]^{1/2} \).

Teacher's Note:
1. An upward accelerating lift increases apparent gravity.
2. Convert the base length into the length along the plane with a cosine.

Answer: (c) 3.69 N

Explanation:
1. Consider the three lower rings (mass 0.3 kg) hanging below the joint.
2. \( T-0.3g=0.3a\Rightarrow T=0.3(9.8+2.5)=3.69\ \text{N} \).

Teacher's Note:
1. Choose the lower portion because it has fewer forces acting on it.
2. Upward acceleration adds to g in the tension.

Answer: (b) \( \left[\frac{2(m_1-m_2)gd}{m_1+m_2}\right]^{1/2} \)

Explanation:
1. Acceleration of the system: \( a=\frac{(m_1-m_2)g}{m_1+m_2} \).
2. \( v^2=2ad\Rightarrow v=\left[\frac{2(m_1-m_2)gd}{m_1+m_2}\right]^{1/2} \).

Teacher's Note:
1. The net driving force is the difference of weights and the moving mass is the sum.
2. The formula v squared = 2ad applies since the acceleration is constant.

Answer: (a) 0.8 m/s²

Explanation:
1. Maximum static friction \( =0.5\times7\times10=35\ \text{N} \). If they moved together, \( a=\frac{100}{42}=2.4\ \text{m/s}^2 \), needing friction \( 35\times2.4=83\ \text{N} \) on A, which is more than 35 N, so B slips on A.
2. Kinetic friction \( =0.4\times7\times10=28\ \text{N} \) acts forward on A.
3. \( a_A=\frac{28}{35}=0.8\ \text{m/s}^2 \).

Teacher's Note:
1. Test whether the blocks can move together by checking the friction required against the maximum static friction.
2. After slipping, use kinetic friction only.

Answer: (b) 5 m/s²

Explanation:
1. Driving force \( =10g-5g\sin30^\circ=100-25=75\ \text{N} \).
2. \( a=\frac{75}{15}=5\ \text{m/s}^2 \).

Teacher's Note:
1. Both masses share the same acceleration since the string is taut.
2. The plane is smooth, so there is no friction term.

Answer: (d) \( g\cot\theta \)

Explanation:
1. For A to stay in contact and fall vertically, the wedge's surface must move away horizontally by x when A falls y with \( y=x\tan\theta \).
2. Differentiating twice: \( g=a\tan\theta \).
3. \( a=g\cot\theta \).

Teacher's Note:
1. Use the geometric constraint between horizontal wedge motion and vertical fall of the body.
2. The body stays on the surface, so the vertical and horizontal displacements are linked by the slope.