A spring (k = 225 N/m) is fixed at the top of a frictionless plane inclined at a
ID: 1705176 • Letter: A
Question
A spring (k = 225 N/m) is fixed at the top of a frictionless plane inclined at an angle ? = 40 degree . A 1.0 kg block is projected up the plane, from an initial position that is distance d = 0.60 m from the end of the relaxed spring, with an initial kinetic energy of 17 J. (a) What is the kinetic energy of the block at the instant it has compressed the spring 0.20 m? (b) With what kinetic energy must the block be projected up the plane (from its initial position) if it is to stop momentarily when it has compressed the spring by 0.40 m?Explanation / Answer
a) At this point, there is only the kinetic energy of the block, which is 17J. Since the spring is compressed .20 m, then the total distance it traveled up the incline is .80 m, including the .60 m. But potential energy only accounts for the vertical distance, so you use x=.80sin(40) to get about .514 m, which you plug into PE = mgh where m is 1 kg, g is 9.8 m/s2 and h is .514, because it traveled up that distance. So the PE should be 5.04 J. But there is also the elastic PE where PE = (.5)kx2 where x is the displacement in only the spring, which means x is .20 m. That PE should be 4.50 J. So the total PE at that moment is 9.54 J. Now just take the original KE (17 J) minus the total PE (9.54 J) and the kinetic energy for part a should be 7.46 J.
b) This time, the kinetic energy is unknown, but we know the gain in Potential Energy and Elastic Potential Energy. PE = mgh + .5kx2. (i just combined the potential energy equations) This time m is 1kg, g is 9.8 m/s2, h is 1 m (.60 + .40), k is 225, and x is .4 m. The PE should be 27.8 J, and since there is no kinetic energy at that point (it has momentarily stopped moving), then the KE required to move the block originally should be 27.8 J. PE=-KE.
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