A 0.111-kg ball is dropped from a height of 1.01 m and lands on a light (approxi
ID: 1985165 • Letter: A
Question
A 0.111-kg ball is dropped from a height of 1.01 m and lands on a light (approximately massless) cup mounted on top of a light, vertical spring initially at its equilibrium position. The maximum compression of the spring is to be 10.5 cm.a) What is the required spring constant of the spring?
b) Suppose you ignore the change in the gravitational energy of the ball during the 10.5-cm compression. What is the percentage difference between the calculated spring constant for this case and the answer obtained in part a)?
Explanation / Answer
Mass m = 0.111 kg Initial height h = 1.01 m Compression x = 10.5 cm = 0.105 m Total height drop H = h + x = 1.115 m From law of conservation of energy , Change in P.E = P.E of the spring mgH = ( 1/ 2) kx^ 2 From thsi spring constant of the spring k = 2mgH / x^ 2 = 220 N / m (b).From law of conservation of energy , Change in P.E = P.E of the spring mgh = ( 1/ 2) k 'x^ 2 From thsi spring constant of the spring k ' = 2mgh / x^ 2 = 199.3 N / m Percentage difference = [( k ' - k ) / k ] x100 = -9.4 %
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