Question

A 65.7 kg skateboarder is sliding down a 15 meter long incline. The incline is 7.50 meters tall. As he slides down he encounters 96.5 N of friction. How fast is the skateboarder moving at the bottom of the hill?

Answer 1

I found `v=10.1m/s`
(by the way, check my maths!)

Explanation:

At the top of the incline he will have gravitational potential energy `U=mgh=65.7*9.8*7.5=4828.9J` that sliding down will be changed into:

Kinetic Energy: `K=1/2mv^2`

and

Heat due to friction equal to the "work" done by friction: `W_f=f*d=96.5*15=1447.5J`

So the Kinetic Energy "available" to move the skater at the bottom of the incline will be the initial potential energy minus the energy used to heat up the ramp:

`K=U-W_f=4828.9-1447.5=3381.4J`
or
`1/2mv^2=3381.4J`
`v=sqrt(2*3381.4/65.7)`

`v=10.1m/s`