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

The correct answer is (C) : 2.0 ms–1
F = –12x
$mv\frac{dv}{dx}=−12$
$∫_{4}^vvdv=−6∫_{0.5}^{1.5}xdx$
(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