Laws of Motion

JEE Main level, single correct, medium

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Question 3 of 210

Two blocks (PP and QQ) with respectively masses 22 kg and 1.51.5 kg are joined by a massless thread. These blocks are mounted on a frictionless pulley which is fixed on the edge of a cube (SS), as shown in the figure below. Block PP is positioned on the top surface which has no friction and block QQ is in contact with side-surface, having coefficient friction μ\mu. The cube (SS) moves towards the right with acceleration of g2\dfrac{g}{2}, where gg is gravitational acceleration. During this movement the block PP and QQ remain stationary. The value of μ\mu is _______. (take g=10g = 10 m/s2^2)

Show answer and solution

Answer: (B) 0.670.67

Let us analyze the forces acting on blocks PP and QQ in the non-inertial frame of reference of the cube SS, which is moving to the right with an acceleration a=g2=5 m/s2a = \dfrac{g}{2} = 5 \text{ m/s}^2.

In this frame, a pseudo force acts on both blocks towards the left.


For block
PP (mass mP=2 kgm_P = 2 \text{ kg}):
The forces in the horizontal direction are the tension
TT in the string (acting towards the right) and the pseudo force FpPF_{pP} (acting towards the left). Since block PP is stationary relative to the cube and there is no friction on the top surface:
T=mPa=2×5=10 NT = m_P a = 2 \times 5 = 10 \text{ N}

For block
QQ (mass mQ=1.5 kgm_Q = 1.5 \text{ kg}):
The horizontal forces are the normal reaction
NN from the side surface of the cube (acting towards the right) and the pseudo force FpQF_{pQ} (acting towards the left). Since block QQ is stationary horizontally relative to the cube:
N=mQa=1.5×5=7.5 NN = m_Q a = 1.5 \times 5 = 7.5 \text{ N}

The vertical forces acting on block
QQ are the downward gravitational force mQgm_Q g, the upward tension TT, and the frictional force ff.
The downward force is
mQg=1.5×10=15 Nm_Q g = 1.5 \times 10 = 15 \text{ N}.
Since
mQg>Tm_Q g \gt T (15 N>10 N15 \text{ N} \gt 10 \text{ N}), the block has a tendency to slide downwards, so the static friction ff must act upwards.
For vertical equilibrium of block
QQ:
T+f=mQgT + f = m_Q g
10+f=15⇒f=5 N10 + f = 15 \Rightarrow f = 5 \text{ N}

To keep the block stationary, the required friction must be less than or equal to the maximum static friction (
f≤μNf \le \mu N). Assuming the limiting case to find the value of μ\mu:
μN=f\mu N = f
μ(7.5)=5\mu (7.5) = 5
μ=57.5=23≈0.67\mu = \dfrac{5}{7.5} = \dfrac{2}{3} \approx 0.67

Answer:
0.670.67

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