Question

A crate whose mass m is 360kg rests on the bed of a truck that is moving at a speed v not of 120km/h.The driver applies the brakes and slows to a speed"v"of 62km/h in 17second .What force (assumed constant) acts on yhe crate during this time ?Assume that the crate does not slide on the truck bed.​

Answer 1

Answer:

As the car rolls down the slope its gravitational potential energy is being converted to kinetic energy and work done against friction.

Loss in gravitational potential energy = gain in Kinetic energy + work done against friction mgh=

2

1

mv

2

+ work done against friction.

⇒ Work done against friction =mgh−

2

1

mv

2

Work done against friction =(500×9.81×30)−(

2

1

×500×11

2

)

Work done against friction =1169005=1.2×10

5

J.

Hence, the answer is 1.2×10

5

J.

Answer 2

Answer:

what doyou mean 17 sec is the answer