Question

After performing a trick above the rim of a skateboard ramp, a 56 kg skateboarder lands on the ramp 3.5 m above ground level with a downward velocity of 4.0 m/s.
Ep = 0.0 J Ground level. Friction in the wheels of the skateboard and air resistance cause a loss of 9.0 × 10^2 J of mechanical energy. The skateboarder’s speed at the bottom of the ramp will be_____m/s.?

Answer 1

potential energy = m g h = 56*9.81 * 3.5
kinetic energy = (1/2) m v^2 = 28*16

sum to get total energy at top

kinetic energy at bottom = (1/2)m v^2
= total energy at top - 900