Question

A box with a mass of 1000 kg is sliding down a ramp with a slope of 15º at an initial speed of 8 m/s.

-Draw a free-body diagram of the car on the hill. All forces must be labelled.

-What coefficient of kinetic friction, k, will result in the acceleration being 0 m/s^2 ?  

-If the coefficient of kinetic friction is actually k=0.4 what is the box acceleration?

-What is the stopping distance?

-After sliding 10 m what is the momentum of thebox?

-At 10m the box collides into another box (m=1000 kg) travelling 3 m/s in the same direction down the slope and they are stuck together. What is the velocity just after impact?

Answer 1

Here, f is friction force

For, acceleration=0

friction force = mg sin∅

If u is coefficient of friction

u mg cos∅= mg sin∅

u=tan∅=tan15°=0.27

If , coefficient of friction=0.4

Acceleration 'a' of block is given by

u mg co15° - mg sin15° = mxa

a= u gcos15° - gsin15°=1.25 m/s²

Acceleration is opposite to the velocity of block

We know,

v=u + at

0=8 - 1.25t

t=6.4s

We know ,

v²=u²+2as

v²=8²-2 x 1.25 x 10

v=6.25m/s

By conservative vation of momentum

m1 x v1 + m2 x v2 =(m1 +m2) x v

v=(1000 x 6.25 + 1000 x 3) /(2000)

=4.6m/s