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A 5-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is suddenly removed. Mkg 2 kg

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Answer #1

K = 40N/m , total weight W = (5+2)×9.8 = 68.6N

Let compression is y.

Hence from Hook's law, W = ky

=) y = W/k = 68.6/40 = 1.715m

Now after removal of 5kg block, m = 2kg

Applying conservation of energy, U = K

=) 1/2 ky^2 = 1/2 mv^2

=) v = y √k/m = 1.715× √20 = 7.67m/s

=) v = 7.67m/s

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