Question

If mass of truck is 3000 kg and it is moving with velocity 90m/s and suddenly applied brakes to come at rest. the brakes applied force of 27000N calculate distance covered by truck and work done by truck​

Answer 1

Answer:

Distance travelled is: 450m

Work done: 12150000J or 12150 KJ

Explanation:

Here,

m=3000kg

F=27000N

v=90m/s

u=0m/s

by F=ma, we get --- a= F/m

so, a= 27000/3000 = 9m/s²

By III equation of motion, v²-u² = 2as

S=450m

also, work= force * displacement

work = 27000*450

         = 12150000J

Answer 2

Answer: Distance=450m

Work done=-12150kJ

Explanation:

Mass=3000kg

Initial Velocity=90m/s

Final Velocity after applying brakes=0m/s

Force=ma=-27000N (Since the brakes are applied to oppose the motion, therefore the force is negative)

3000×a=-27000

a=-9m/s^2

We know that v^2-u^2=2aS

S=-8100/(2×(-9))

S=450m

We know that We=FS cos (theta)

And since the breaks are opposing the motion, i.e, the angle between force and displacement is 180°, therefore the value of cos(theta) will be -1

W= 27000×450×(-1)

=-12150000J =-12150kJ

Hope my answer helped :)