Asked by Julius
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.
Answers
Answered by
drwls
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
(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
Answered by
Julius
thank you
Answered by
Anonymous
where did you get 1.791 from
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