Problem Set 4 - 5.32) Cart in a Valley The cart’s initial...

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Unformatted text preview: 5.32) Cart in a Valley The cart’s initial potential energy is mgh 1 There is no initial kinetic energy (cart is said to be at rest), so this is the total energy of the cart. As a quantity δm of sand leaks out of the cart, the cart’s energy will be reduced by δmgy ( t )+ δmv ( t ) 2 / 2, where v and y are the current speed and height of the cart. However, we see that the proportionality between the cart’s mass and its energy remains constant. So E 1 m 1 = E 2 m 2 → gh 1 = gh 2 and the cart will reach the same height on the other side of the valley. 5.33) Walking on an Escalator Work is force times displacement. If you remain at rest with respect to the ground frame (displacement is zero), then you are doing no work in that frame. 5.35) Spring Energy Given a spring moving with x ( t ) = A cos( ωt + φ ) we can compute kinetic energy K = 1 2 m ˙ x 2 = m 2 (- ωA sin( ωt + φ ) 2 = mω 2 A 2 2 sin 2 ( ωt + φ ) and potential energy U = 1 2 kx 2 = k 2 A 2 cos 2 ( ωt + phi ) Sum these two quantities to find the total energy E = k 2 A 2 cos 2 ( ωt + φ ) + mω 2 A 2 2 sin 2 ( ωt + φ ) and use the fact that ω 2 = k m to see E = k 2 A 2 cos 2 ( ωt + φ )+ kA 2 2 sin 2 ( ωt + φ ) = kA 2 2 (cos 2 ( ωt + φ )+sin 2 ( ωt + φ )) = kA 2...
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This note was uploaded on 01/22/2012 for the course PHYS 3201 taught by Professor Deheer during the Spring '10 term at Georgia Institute of Technology.

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Problem Set 4 - 5.32) Cart in a Valley The cart’s initial...

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