Question

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 to 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
• D. 2.5 ms–1

Answer 1

Answer: (C)

2.0 ms–1

Answer 2

Given: $k=12 Nm^{-1}$
$F = –kx$
$F = –12x$

$mv \text{} \frac{dv}{dx} =−12$

$\int_{4}^{v}vdx=-6\int_{0.5}^{1.5}xdx$
(m = 2 kg)

$\frac{v^2-16}{2}$=−6[$\frac{1.5^2-0.5^2}{2}$]

$\frac{v^2 -16}{2}$=−6
v = 2 m/sec

$\therefore ,$ The correct option is (C): $2.0\text{ ms}^{-1}$