A 448 g block is released from rest at height h 0 above a vertical spring with s

Question

A 448 g block is released from rest at height h0 above a vertical spring with spring constant k = 340 N/m and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 17.4 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 4h0 above the spring, what would be the maximum compression of the spring?

(a) Number Units This answer has no units° (degrees)mkgsm/sm/s^2NJWN/mkg·m/s or N·sN/m^2 or Pakg/m^3gm/s^3times

Explanation / Answer

given

m = 448 g = 0.448 kg

k = 340 N/m

y = 17 4 cm = 0.174 m

a) workdone by block on spring = (1/2)*k*y^2

= 0.5*340*0.174^2

= 5.147 J

b) workdone by spring on block = -5.147 J

c) Apply conservation of energy

initial gravitational PE = final elastuc PE

m*g*(ho + y) = (1/2)*k*y^2

ho + y = (1/2)*k*y^2/(m*g)

ho = (1/2)*k*y^2/(m*g) - y

= 5.147/(0.448*9.8) - 0.174

= 0.998 m or 99.8 cm

d)

Again apply conservation of energy

initial gravitational PE = final elastuc PE

m*g*(4*ho + y) = (1/2)*k*y^2

4*m*g*ho + m*g*y = 0.5*k*y^2

4*0.448*9.8*0.998 + 0.448*9.8*y = 0.5*340*y^2

170*y^2 - 4.39*y - 17.53 = 0

on solving the above equation we get

y = 0.334 m or 33.4 cm

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