A block of mass 2 kg moving on a horizontal surface with speed of $4 \mathrm{~ms}^{-1}$ enters a rough surface ranging from $\mathrm{x}=0.5 \mathrm{~m}$ to $\mathrm{x}=1.5 \mathrm{~m}$. The retarding force in this range of rough surface is related to distance by $\mathrm{F}=-\mathrm{kx}$ where $\mathrm{k}=12 \mathrm{Nm}^{-1}$. The speed of the block as it just crosses the rough surface will be:

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Work, Power & Energy

A block of mass 2 kg moving on a horizontal surface with speed of

4 ms^{- 1}

enters a rough surface ranging from

x = 0.5 m

x = 1.5 m

. The retarding force in this range of rough surface is related to distance by

F = - kx

where

k = 12 Nm^{- 1}

. The speed of the block as it just crosses the rough surface will be:

(A)Zero
(B) $1.5 ms^{- 1}$
(C) $2.0 ms^{- 1}$ ✓ Correct
(D) $2.5 ms^{- 1}$

Answer: (C)

Solution

Using

F = ma = - k x

, we get

a = - \frac{k x}{m} = - 6 x

. Using

v \frac{d v}{d x} = a

\int_{4}^{v} v d v = - \int_{0.5}^{1.5} 6 x d x

. This gives

\frac{v ^{2} - 16}{2} = - 3 [(1.5)^{2} - (0.5)^{2}] = - 3 (2) = - 6

. Therefore

v^{2} = 4

, so

v = 2

m/s.

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option

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