JEE Physics 2023 Question with Solution

Given: Mass of the body is $10 \, \text{kg}$, initial speed is $20 \, \text{m/s}$, final speed is $0$, time taken is $5 \, \text{s}$, and $g = 10 \, \text{m/s}^2$.

Find: The coefficient of friction $\mu$.

The solution uses retardation due to friction:

$$a = -\mu g$$

Also, from the equation of motion:

$$v = u + at$$

Substituting the given values:

$$0 = 20 + (-\mu \times 10) \times 5$$

So,

$$0 = 20 - 50\mu$$ $$50\mu = 20$$ $$\mu = \frac{2}{5}$$

Hence,

$$\mu = 0.4$$

Therefore, the value of coefficient of friction is $0.4$. The correct option is D.

The solution labels the option as B, but its worked value is $0.4$, which matches option D in the given options.

Given: Mass $m = 10 \, \text{kg}$, initial speed $v_i = 20 \, \text{m/s}$, final speed $v_f = 0$, time $t = 5 \, \text{s}$, and $g = 10 \, \text{m/s}^2$.

Find: The coefficient of friction $\mu$.

The work done by the frictional force is equal to the change in kinetic energy of the body.

First, find the acceleration using:

$$v_f = v_i + at$$

Substituting values:

$$0 = 20 + a \times 5$$ $$a = -4 \, \text{m/s}^2$$

Using Newton's second law:

$$F = ma$$ $$F = 10 \times (-4) = -40 \, \text{N}$$

The magnitude of frictional force is $40 \, \text{N}$.

Now,

$$F = \mu mg$$

So,

$$40 = \mu \times 10 \times 10$$ $$\mu = \frac{40}{100} = 0.4$$

Therefore, the coefficient of friction is $0.4$ and the correct option is D.