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?

Nov 23, 2015

I found $v = 10.1 \frac{m}{s}$
(by the way, check my maths!)

Explanation:

At the top of the incline he will have gravitational potential energy $U = m g h = 65.7 \cdot 9.8 \cdot 7.5 = 4828.9 J$ that sliding down will be changed into:

Kinetic Energy: $K = \frac{1}{2} m {v}^{2}$

and

Heat due to friction equal to the "work" done by friction: ${W}_{f} = f \cdot d = 96.5 \cdot 15 = 1447.5 J$

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:

$K = U - {W}_{f} = 4828.9 - 1447.5 = 3381.4 J$
or
$\frac{1}{2} m {v}^{2} = 3381.4 J$
$v = \sqrt{2 \cdot \frac{3381.4}{65.7}}$

$v = 10.1 \frac{m}{s}$