Question

A body starting from rest slides over a plane inclined at 45° to the horizontal. After certain displacement, body has velocity 2m/s. If coefficient of friction between body and surface is 0.5, find displacement of the body from a start.

Answer 1

Let mass of body = m
inclination of plane , $\theta$ = 45°
coefficient of friction, u = 0.5

initial velocity of body , u = 0
final velocity of body , v = 2m/s

use Newton's 2nd law,
Net force = mgsin$\theta$ - umgcos$\theta$
ma = mg(sin$\theta$ -ucos$\theta$)
a = 10(sin45° - 0.5cos45°)
a = 10(1/√2 - 1/2√2) = 10/2√2 = 5/√2 m/s²

now, use formula, v² = u² + 2as
2² = 0² + 2 × 5/√2 × s
4 = 5√2s
s = 4/5√2 = 2√2/5 = 2.828/5 = 0.5656m