A spring has a relaxed length of 30 cm (0.30 m) and its spring stiffness is 12 N

Question

A spring has a relaxed length of 30 cm (0.30 m) and its spring stiffness is 12 N/m. You glue a 66 gram block (0.066 kg) to the top of the spring, and push the block down, compressing the spring so its total length is 13 cm. You make sure the block is at rest, then at time t = 0 you quickly move your hand away. The block begins to move upward, because the upward force on the block by the spring is greater than the downward force on the block by the Earth. Calculate y vs. time for the block during a 0.27-second interval after you release the block, by applying the Momentum Principle in three steps each of 0.09-second duration.

Explanation / Answer

F(on block)=(1/2)ky^2(the y^2 term is in reference to the total deviation from relaxed. So, then, relaxed length-compressed length yields y.) F(gravity on block)=mg F(net) =F(on block)+F(gravity) dont forget that the force due to gravity is negative. a=F(net)/m; v(final)=v(original)+(acceleration)x(time); p=mv (change in y)=y(original)+v(original)+(1/2)at^2 if you repeat this for each .09 second interval, you will be able to create your graph.

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