Rollercoaster mechanics question

A rollercoaster at a fairground has a mass of 380kg.
The coaster is pulled up to point P which is 20m high.
The rollercoaster at point P is travelling at 2ms^-1.
At the point P, the car is 20m above horizontal ground and is travelling at 2ms^-1. It then travels 100m along the track in 20 seconds to a point Q that is 5m above ground. At Q the cart is travelling at 10ms^-1.

i) Show the change in KE of the cart from P to Q
ii) Show the change in GPE of the car, stating whether it is a gain or loss.

During motion from P to Q, the average braking force over the distance is 150N.

iii) Show that the total work done against the resistances to motion other than braking is 22620J.

At point Q, the cart goes up a uniform slope at 20 degrees to the horizontal and comes instantaneously to rest at the point R. The average resistances to motion and the average braking force from Q to R are the same as in the motion from P to Q.

iv) Calculate the vertical distance of R above Q.

I'm fine with parts i) and ii) it just gets a bit ugly after that.

From i) and ii) you can work out the loss in energy.

This was caused by two things, work done against braking, and work done against resistances to motion. You can calculate the former, and hence determine the latter.

From i) and ii) you can work out the loss in energy.

This was caused by two things, work done against braking, and work done against resistances to motion. You can calculate the former, and hence determine the latter.

I did do that but still came out with the wrong answer. I'm probably wrong so i'll just check my workings.

Using g=9.8 m/s/s it does work out to the desired figure.

Could I just ask what your GPE and KE figures were?

I didn't work out the intermediate results, but:

Loss in GPE $= (20-5)\times 9.8\times 380$

Gain in KE $= \dfrac{380}{2}(10^2-2^2)$

And work done against braking $= 150\times 100$