Hey guys!
I am doing my physics assignment and theres a question that I don't fully know how to do...
Heres the question!

A 50.0kg toboggan is moving at a speed of 2.2 m/s when it crests a hill. The height of the hill is 10.0m. How fast will the sled be moving when it passes a slower sled 2.0 m from the bottom of the hill ?

Here's what I did...

I got the Total Energy (ET) from when the sled was at the top of the hill, and I also got the Total Energy (ET) when the sled is 2.0 m from the bottom of the hill..

My physics teacher told me the next step was to subtract the two values.. so i did as follows
5021 J - 1101 J = 3920 J.

Im not quite sure what to do from here...
I thought of doing ET = mgh + 1/2mv2.. but im not sure how to go about rearranging that to find the v2 because theres 2 masses on the right side of the equation..

anyone know what to do or know if i am doing it wrong?
thanks !!

By conservation of energy,
total energy before = total energy after

At the top of the hill, K.E. =0 and P.E. = 5021J
At 2m above the bottom, some of the potential energy is changed to kinetic energy, K.E. = (1/2)(m)(v^2) P.E. = mgh=50(9.8)(2)

Therefore,
5021=(1/2)(mv^2) +2010
5021-2010=(1/2)(50)(v^2)
Then solve for v

__________________
Going to face the Hong Kong Advanced-Level Examination which is believed to be the most difficult AL exams in the world this year. I hope good results can be achieved.