Posted by **Julius ** on Saturday, February 19, 2011 at 2:37pm.

A student is skateboarding down a ramp that is 5.5 m long and inclined at 19° with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is 2.9 m/s. Neglect friction and find the speed at the bottom of the ramp.

- physics -
**drwls**, Saturday, February 19, 2011 at 4:47pm
Apply conservation of energy. Neglect the rotational kinetic energy of the wheels, which will make a small difference since the wheels weigh much less than the skateboarder.

(M/2)[V2^2 - V1^2] = M g deltaH

= M g *5.5 sin 19

V1 = 2.9 m/s

Cancel the M's and solve for V2, the final velocity.

V2^2 - (2.9)^2 = 2*9.81*1.791 = 35.13

V2^2 = 35.13 + 8.41

V2 = 6.60 m/s

- physics -
**Julius **, Saturday, February 19, 2011 at 4:51pm
thank you

## Answer this Question

## Related Questions

- Physics - A student is skateboarding down a ramp that is 6.03 m long and ...
- physics - A student is skateboarding down a ramp that is 7.06 m long and ...
- Physics - A student is skateboarding down a ramp that is 8.33 m long and ...
- Physics - A student is skateboarding down a ramp that is 8.81 m long and ...
- physics - A student is skateboarding down a ramp that is 8.48 m long and ...
- physics - A student is skateboarding down a ramp that is 6.52 m long and ...
- physics - A student is skateboarding down a ramp that is 6.88 m long and ...
- Physics - A student is skateboarding down a ramp that is 5.38 m long and ...
- Physics - Consider a skateboarder who starts from rest at the top of a 14.6 m-...
- Physics - Consider a skateboarder who starts from rest at the top of a 12.5m-...

More Related Questions