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There is a ramp with height "h". There is a box at the top of the ramp and it starts at rest, v0=0. There is no friction on the ramp. The catch is that this ramp is in a moving truck, moving at velocity "u". Once the box is released, it moves at it's own velocity "v" plus the velocity of the truck "u", if I am correct. So the total kinetic energy of the box should be: K=(1/2)mu^2+(1/2)mv^2 or (1/2)m(u^2+v^2). However, I looked at the back of the book and the answer says (1/2)m(u+v)^2, which is not equal to the other equation. What did I do wrong? If you need further explanation please ask me. I am in dire need of help!

2007-11-07 10:30:47 · 1 answers · asked by texasrag31 1 in Science & Mathematics Physics

1 answers

Total kinetic energy is equal to 1/2 m V^2 where V is the total velocity of the box. After being released, the total velocity of the box is V = u + v so its kinetic energy must be 1/2 m (u+v)^2, which is equal to 1/2 m(u^2 + 2uv + v^2).

2007-11-07 10:52:16 · answer #1 · answered by msi_cord 7 · 0 0

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