Question

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.

Answer 1

Step-by-step explanation:

Given

inclination
`\theta =12^(\circ)`

Initial velocity of rock
`u=15\ m/s`

length of track
`s=2400\ m`

Theoretical speed of rock can be given by using

`v^2-u^2=2as`

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

here gravity sin component will accelerate the rock

`a=g\sin 12`

`v^2-(15)^2=2* (g\sin (12))\cdot 2400`

`v^2=225+9780.165`

`v=√(10,005.165)`

`v=100.025\ m/s`

It is given that velocity at the bottom is
`v=40\ m/s`

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

`W_g+W_f=\Delta k`

where
`W_g`=gravity work

`W_f`=Friction work

`\Delta k`=change in kinetic Energy

`W_g=mgh=m\cdot g\cdot 2400\sin 12`

`W_g=mg\cdot =498.98mg`

`W_f=f\cdot 2400`

`\Delta k=0.5\cdot m(100.02^2-40^2)`

substituting values we get

`f_(avg)=1.75 m`

where m is the mass of rock