Question

A car of 600 kg travelling with a velocity of 72 km/h is stopped in a distance of 300m find the breaking force

Answer 1

Given,

Initial velocity, u=72 km/h

u=360072×1000 m/s=20 m/s

Final velocity, v=0 m/s

Displacement s=60 m

Using Newton's third equation of motion,

2as=v2−u2

⟹a=2sv2−u2

⟹a=2×6002−202

⟹a=2×6002−202

⟹a=120400

a=−310 m/s2

Magnitude of acceleration, a=3−10 m/s2

Force F=ma=600×(10/3)

F=2000 N

Work done, W=F×S

W=2000×60

W=120000J