A 700 g block is released from rest at height h 0 above a vertical spring with s

Question

A 700 g block is released from rest at heighth0 above a vertical spring with spring constantk = 345 N/m and negligiblemass. The block sticks to the spring and momentarily stops aftercompressing the spring 22.5 cm.


(a) How much work is done by the block on thespring?
1
(b) How much work is done by the spring on the block?

(c) What is the value of h0?

(d) If the block were released from height2.00h0 above the spring, what would be themaximum compression of the spring?
(a) How much work is done by the block on thespring?
1
(b) How much work is done by the spring on the block?

(c) What is the value of h0?

(d) If the block were released from height2.00h0 above the spring, what would be themaximum compression of the spring?

Explanation / Answer

(a) W on the spring by the block = Loss of its Grav PE =mgh'              Wsb = 0.70x9.8(ho +0.225) = 6.86(ho +0.225)J (b) Work done by the spring on the block, Wbs = -Wsb assuming no loss as heat etc.             Wbs= - Wsb = - 6.86(ho +0.225)J (c) ho =? Energy stored in the spring,            Es = 0.5 k.x2 = 8.859 J Equating this to W done on the spring               ho = 1.291 - 0.225 = 1.066 m Going back using this ho, (a)   Wsb= - Wsb=  8.86 J (b)   Wbs = - Wsb = -8.86 J Added later (d) When dropped from a height of 2ho =2x1.066 = 2.132 m Let y be the compression; total height, h' = 2.132 + y Loss of PE, U = mgh' = 0.7x9.8x (2.132 +y) = 6.86(2.132 +y) Energy stored in the spring, Es =0.5k.y2 = 0.5x350y2 = 175.y2 Equating, U = Es                 175.y2 = 6.86 (2.132 +y)                  25.51 y2 - y - 2.132 = 0                       y = 0.309m

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