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# A 884 g block is released from rest at height h0 above a vertical spring with spring constant k = 570 N/m and negligible mass.?

A 884 g block is released from rest at height h0 above a vertical spring with spring constant k = 570 N/m and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 14.0 cm. How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of h0? (d) If the block were released from height 6h0 above the spring, what would be the maximum compression of the spring?

### 1 Answer

- WoodsmanLv 72 years agoFavorite Answer
(a) the work done on the spring is

W = SPE = ½kx² = ½ * 570N/m * (0.140m)² = 5.59 J

(b) BY the spring, -5.59 J since it is against the block's direction of motion

(c) GPE got converted into SPE:

mg(h0 + x) = 5.59 J

0.884kg * 9.8m/s² * (h0 + 0.140m) = 5.59 J

h0 = 0.505 m

(d) 0.884kg * 9.8m/s² * (6*0.505m + x) = ½ * 570N/m * x²

has roots at x = -0.289 m ← ignore

and x = 0.319 m

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