A block of mass 1 kg, moving along x with speed $v_{i}=10\,\text{m/s}$ enters a rough region ranging from $x=0.1\,\text{m}$ to $x=1.9\,\text{m}$. The retarding force acting on the block in this range is $F_{r}=-kx\,\text{N}$, with $k=10\,\text{N/m}$. Then the final speed of the block as it crosses rough region is

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Work, Power & Energy

A block of mass 1 kg, moving along x with speed

v_{i} = 10 m/s

enters a rough region ranging from

x = 0.1 m

x = 1.9 m

. The retarding force acting on the block in this range is

F_{r} = - k x N

, with

k = 10 N/m

. Then the final speed of the block as it crosses rough region is

(A) $10 m/s$
(B) $4 m/s$
(C) $6 m/s$
(D) $8 m/s$ ✓ Correct

Answer: (D)

Solution

The work done by the retarding force as the block crosses the rough stretch is

\int_{0.1}^{1.9} (- k x) d x = - \frac{k}{2} (1. 9^{2} - 0. 1^{2}) = - \frac{10}{2} (3.6) = - 18 J

. By the work–energy theorem,

\frac{1}{2} m v_{f}^{2} = \frac{1}{2} (1) (10)^{2} - 18 = 32 J

, so

v_{f}^{2} = 64

. The block leaves the rough region at

8 m/s

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Single correct option

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