attn mechanical engineers or physic gurus, page-14

There was another question related to this, however I thought it wasn't worth posting because it was asking for the force action on the trolley... here it is

A system of a 120kg empty trolley and a 100kg counterweight is as shown (refer pic)



After the counterweight has dropped 5m from rest, the velocity of the trolley is 3m/s. If all the frictional resistances can be regarded as a single force F action on the trolley parallel to the plane, calculate the resistance force.

Solution
the trolley will move 2x the velocity (v) of the counterweight and 2x the distance (S)
(PE1+KE1)+-W=(PE2+KE2+)

Trolley S=10m v=3m/s
(0 + 0)+-10F=(120x3^2 x1/2 + 120x9.81x10xSIN20)

Counterweight S=5m v=1.5m/s
(0 + 100x9.81x5)+-0=(100x1.5^2 + 0)

Result
(0+4905)-10F=(540+112.5+4026.26)
F=226.24/10
F=22.6

Thanks for your input guys. It's appreciated!

Could the both potiental energy be cancelled out due to unknown height?