Question

A 3000kg truck moving at a speed of 90m per sec stops after covering some distance.The force applied by brakes is 27000N.Compute the distance covered and work done by this force.

Answer 1

Hey!

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Force = Mass × acceleration

Given,
Mass= 3000 kg
Initial velocity = 90m/s
Final velocity= 0 m/s
Force= 27000 N

Acceleration=?
Force/Mass= Acceleration
27000/3000= Acceleration

Acceleration= -9m/s^2 [ '-' is used because speed was reduced]

Distance covered=?
Through third law of motion, we can find the answer.

2as= v^2 - u^2
s= v^2-u^2/2a
s= 0-90^2/2×-9
s= 450 m

Distance covered= 450 m

Work done by the force= F×s
Work done= 27000×450
Work done= 12150000 J
Work done= 12150 kJ

Work done by the force = 12150 kJ

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Hope it helps...!!!