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| Math Central | Quandaries & Queries |
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Question from KOBINA, a student: A 50 kg block being held at rest 20m above the ground is released. the block falls (no friction). how fast is the block traveling (in m/sec) when it has lost 40% of its original potential energy. Question |
Hi Kobina.
The formula for potential energy is simply PE = mgh. When 40% of it has been lost, then that energy has become kinetic energy, so KE = (0.40)mgh.
But KE is related to the speed of the object and its mass: KE = (1/2) mv2. We're interested in the velocity, not the amount of KE or PE, so we can set these two expressions equal to each other:
(0.40)mgh = (1/2)mv2
And when we solve for the velocity, we notice that mass doesn't event matter, thus
v = sqrt(0.8(gh))
g is a constant: 9.8 m/s2 and h is given: 20 m. So you can complete the calculation and find the answer.
Cheers,
Stephen La Rocque
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