You are towing a trailer of large rocks up a steep road that is inclined 120 to the horizontal. Although you are careful to drive at a steady speed of 15 m/s, when you get 2400 m along the road one of the rocks slides off the back of the trailer and rolls down the incline. How fast is the rock going when it reaches the bottom? If you find instead that the speed of the rock is 40 m/s at the bottom, what is the average drag force on the rock? Assume that the trailer is low enough to the ground that the rock does not lose any energy as it hits the ground.

Explanation:

Given

inclination

Initial velocity of rock

length of track

Theoretical speed of rock can be given by  using

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

here gravity sin component will accelerate the rock

It is given that velocity at the bottom is

this is because of friction

According to work-energy theory Change in kinetic energy of  an object is given by work done by all the forces

where =gravity work

=Friction work

=change in kinetic Energy

substituting values we get

where m is the mass of rock                          

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards