The figure 8-41 shows an 4.09 kg stone at rest on a spring. Thespring is compres

Question

The figure 8-41 shows an 4.09 kg stone at rest on a spring. Thespring is compressed 9.62 cm by the stone. (a)What is the spring constant? (b) The stone ispushed down an additional 34.0 cm and released. What is the elasticpotential energy of the compressed spring just before that release?(c) What is the change in the gravitationalpotential energy of the stone-Earth system when the stone movesfrom the release point to its maximum height? (d)What is the maximum height, measured from the release point?

Explanation / Answer

(a) spring constant = mg / x = 4.09 *9.80 / 0.0962 =   416.65 N/m . (b)   elastic PE in spring = (1/2) kx2 = (1/2) * 416.65 * (0.34 +0.0962)2=    39.64Joules . (c)   Since energy is conserved, and the finalenergy stored in the spring is zero, the gravitational potentialenergy of the stone-Earth system increases by  39.64 Joules . (d)    PE = m g h .        h = PE/ mg = 39.64 / 4.09 * 9.80 = 0.989 meters = 98.9cm

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