# 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...

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 x 10^2 J of mechanical energy.

The skateboarder’s speed at the bottom of the ramp will be____m/s.?

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### 1 Answer

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. The reference energy at ground level is 0.

Friction in the wheels of the skateboard and air resistance cause a loss of 9.0*10^2 J of mechanical energy.

As the skateboarder slides down the ramp the potential energy of the skateboarder is converted to kinetic energy. Potential energy at a height of 3.5 m is equal to 56*9.8*3.5 = 1920.8 J. There is a loss of 9*10^2 = 900 J due to friction and air resistance. The net change in mechanical energy of the skateboarder at the bottom of the ramp compared to that at the top is 1020.8 J.

The kinetic energy of an object with mass m, moving at v m/s is given by the formula ME = `(1/2)*m*v^2` .

This gives the increase in velocity as `sqrt((1020.8*2)/56) ~~ 6.038` m/s

The skateboarder had an initial downward speed of approximately 6 m/s when he landed on the ramp. **The speed of the skateboarder at the bottom of the ramp is 10.038 m/s.**