Question

A Car of 600kg travelling with a velocity of 72km/h 15 stopped in a distance of 300m. Find the breaking force

Answer 1

Answer:

1.2×105 J

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×

Answer 2

1.2×10 5J

Given,

Initial velocity, u=72 km/h

u= 3600

72×1000 m/s=20 m/s

Final velocity, v=0 m/s

Displacement s=60 m

Using Newton's third equation of motion,

2as=v

2 −u 2⟹a= 2s=v −u 2⟹a= 2×600 2× −20 2

⟹a=2×60

2× −20 2

⟹a= 120×400

a=− 3×10 m/s 2

Magnitude of acceleration, a= 3×−10 m/s 2

Force F=ma=600×(10/3)

F=2000 N

Work done, W=F×S

W=2000×60

W=120000J