Question

A block of mass 40 kg slides over a surface, when a mass of 4 kg is suspended through an inextensible massless string passing over a frictionless pulley as shown below.
The coefficient of kinetic friction between the surface and block is 0.02. The acceleration of the block is (Given g = 10 ms^–2)

• A. 1 ms^–2
• B. \(\frac{1}{5}\) ms^–2
• C. \(\frac{4}{5}\) ms^–2
• D. \(\frac{8}{11}\) ms^–2

Answer 1

The Correct Option is D

To find the acceleration of the block, we need to analyze the forces acting on both the 40 kg block on the surface and the 4 kg block hanging from the pulley.

Let's denote:

• \( M = 40 \, \text{kg} \): mass of the block on the surface
• \( m = 4 \, \text{kg} \): mass of the hanging block
• \( g = 10 \, \text{ms}^{-2} \): acceleration due to gravity
• \( \mu = 0.02 \): coefficient of kinetic friction
• \( T \): tension in the string
• \( a \): acceleration of the system

For the 40 kg block on the surface, the forces are:

• Tension in the string (\( T \))
• Kinetic friction (\( f_k = \mu \cdot M \cdot g \))

The net force equation for the 40 kg block is:

T - \mu \cdot M \cdot g = M \cdot a

For the 4 kg hanging block, the forces are:

• Gravitational force (\( m \cdot g \))
• Tension in the string (\( T \))

The net force equation for the 4 kg block is:

m \cdot g - T = m \cdot a

Substituting the known values:

1. For the 40 kg block: T - 0.02 \cdot 40 \cdot 10 = 40 \cdot a
2. For the 4 kg block: 4 \cdot 10 - T = 4 \cdot a

Simplifying the equations:

1. T - 8 = 40a
2. 40 - T = 4a

Solving these two equations simultaneously:

• Adding the two equations:

  40 - 8 = 44 a \Rightarrow 32 = 44a \Rightarrow a = \frac{32}{44} = \frac{8}{11} \, \text{ms}^{-2}

Therefore, the acceleration of the block is \frac{8}{11} \, \text{ms}^{-2}.