A skateboarder with a mass of 82.9kg is on top of a hill of height 2.8m. What will be the final velocity at the bottom if no energy is lost due to friction?

Question

A skateboarder with a mass of 82.9kg is on top of a hill of height 2.8m. What will be the final velocity at the bottom if no energy is lost due to friction?

Answer 1

Hi Chloe B.,

Just set up your Conservation of Energy equation:

Gravitational Potential Energy = Kinetic Energy gained

mgh =(1/2) mv^2

Now, you may cancel the mass on each side (identical terms) to get"

gh = (1/2)v^2

v^2 = 2gh

Next, plug in your knowns: (you should memorize the value for g)

=2*(9.8m s^-2)(2.8 m)

Solve on through, then take the square root.

-- Cheers, -- Mr. d.

Answer 2

If no energy is lost due to friction, you can apply the principle of conservation of energy.

At the top of the hill, the kinetic energy KE is zero since the skateboard is at rest. It's potential energy is given by PE=mgh where m is the mass of the skateboard, g is the gravitational acceleration, and h is the height of the hill. Therefore the total energy TE is given by TE=KE+PE=mgh.

At the bottom of the hill, the kinetic energy KE will be KE=0.5mv² and the potential energy will be PE=0 since you are at height equal zero. The total energy at the bottom of the hill is given by TE=KE+PE=0.5mv²

The total energy must be conserved and hence, mgh=0.5mv² which implies v=(2gh)^1/2

With the numbers you have v=(2x2.8x82.9)^1/2≈21.54 m/s

Best,

Davide