JEE Physics 2023 Question with Solution

Given: $m = 5 \, \text{kg}$, applied force $F = 30 \, \text{N}$, displacement $S = 50 \, \text{m}$, time $t = 10 \, \text{s}$, initial velocity $u = 0$, and $g = 10 \, \text{m s}^{-2}$.

Find: The coefficient of kinetic friction $\mu$.

Using the kinematic equation,

Substituting the given values,

So,

Now apply Newton's second law along the surface:

Substituting the values,

Hence,

Therefore, the coefficient of kinetic friction is $0.50$. The solution working gives $\mu = \frac{1}{2}$, which corresponds to option C. However, the solution states "The Correct Option is A," which is inconsistent with the working.

Given: $S = 50 \, \text{m}$, $t = 10 \, \text{s}$, $u = 0$, $m = 5 \, \text{kg}$, $F = 30 \, \text{N}$, $g = 10 \, \text{m s}^{-2}$.

Find: $\mu$.

First find the acceleration from motion data:

The normal reaction is

So the kinetic friction is

Net force along the direction of motion is

Thus, the correct value is $0.50$, so the defensible correct option is C.