A 0.500-kg block attached to a spring with length 0.60 m and force constant 40.0

Question

A 0.500-kg block attached to a spring with length 0.60 m and force constant 40.0 N/m is at rest with the back of the block at point A on a frictionless, horizontal air table (see figure below). The mass of the spring is negligible. You pull the block to the right along the surface by pulling with a constant F = 20.8-N horizontal force. (a) What is the block's speed when the back of the block reaches point B, which is d = 0.20 m to the right of point A? (b) When the back of the block reaches point B, you let go of the block. In the subsequent motion, how close does the block get to the wall where the left end of the spring is attached?

Explanation / Answer

a) From work energy theorem

work done = F * d = 1/2 m v2 + 1/2 k e2

(20.8 * 0.20) = (1/2 * 0.500 * v2) + (0.5 * 40 * 0.202)

v = 3.67 m/s

b) W = 20.8 * 0.20 = 4.16 J

1/2 k d2 = 4.16

d2 = 2 * 4.16 / 40

d = 0.456 m

distance from wall = 0.60m - 0.456 = 0.144 m

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