A smooth plane AB is 10m long. It is inclined at 30 degrees to the horizontal with the lower end, B 6m vertically above horizontal ground. A particle is placed on the plane at the upper end, A, and then released from rest so that it slides down the plane. Find the speed of the particle as it strikes the ground.
My initial idea was to apply mgh = 1/2mv^2 since there aren't any resistive forces to get v = sqrt (2gh) = sqrt (2x9.8x6) = 10.8ms^-1
This didn't work so I tried using suvat with a = gsin30 = 4.9ms^-2. s = 10m and u = 0ms^-1. Then v = sqrt (0^2 +2 x 4.9 x 10) = 9.90ms^-1
The correct answer is 14.7ms^-1. If anyone could point out where I have gone wrong in each method and how to get the correct answer it would be appreciated!!
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- Thread Starter
- 10-04-2011 11:13
- 10-04-2011 11:19
Draw a diagram!
It slides down the plane and then falls another 6m.
- 10-04-2011 11:24
At the top it only has PE as it is released from rest.
At the bottom it will only have KE, so all PE -> KE.
So you need to work out what is the PE at the top. Its height will be 6m + the height of the slope which can be worked out with a bit of basic trig.