JEE Physics 2024 Question with Solution

Given: Mass of the block is $1 \, \text{kg}$, angle of inclination is $60^\circ$, distance moved along the incline is $10 \, \text{m}$, coefficient of kinetic friction is $0.1$, and take $g = 10 \, \text{m/s}^2$.

Find: Work done against the frictional force.

The kinetic frictional force is

$$f_k = \mu N$$

For a block on an inclined plane, the normal force is

$$N = mg\cos\theta$$

Substituting the values,

$$N = 1 \times 10 \times \cos 60^\circ = 1 \times 10 \times \frac{1}{2} = 5 \, \text{N}$$

Now the frictional force is

$$f_k = 0.1 \times 5 = 0.5 \, \text{N}$$

The work done against friction is

$$W = f_k \times d$$

So,

$$W = 0.5 \times 10 = 5 \, \text{J}$$

Therefore, the work done against the frictional force is $5 \, \text{J}$.

Given: $m = 1 \, \text{kg}$, $\theta = 60^\circ$, $\mu = 0.1$, $d = 10 \, \text{m}$, $g = 10 \, \text{m/s}^2$.

Find: The work done against friction.

First calculate the normal reaction on the block:

$$N = mg\cos\theta$$ $$N = (1)(10)\cos 60^\circ$$ $$N = 10 \times 0.5 = 5 \, \text{N}$$

Now calculate the kinetic frictional force:

$$f_k = \mu N$$ $$f_k = 0.1 \times 5 = 0.5 \, \text{N}$$

The work done against friction over distance $10 \, \text{m}$ is:

$$W_{\text{against friction}} = f_k \cdot d$$ $$W_{\text{against friction}} = 0.5 \times 10$$ $$W_{\text{against friction}} = 5 \, \text{J}$$

Hence, the required numerical value is $5$.