A 0.50 kg mass sliding on a horizontal frictionless surface is attached to one e
ID: 1331301 • Letter: A
Question
A 0.50 kg mass sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (with k = 450 N/m) whose other end is fixed. The mass has a kinetic energy of 17.0 J as it passes through its equilibrium position (the point at which the spring force is zero). At what rate is the spring doing work on the mass as the mass passes through its equilibrium position?
0.00 W
At what rate is the spring doing work on the mass when the spring is compressed 0.137 m and the mass is moving away from the equilibrium position?
Explanation / Answer
Let W is frequency of spring
work done by spring = 1/2*K*X^2
rate of work done = KXdX/dt = K*X*V
a) at equilibrium position ( X=0)
rate of work done = K*X*V = 0
K.E at equilibrium = 1/2*A*W^2 = 17
W = sqrt (k/m) = sqrt (450/0.50) = 30
V = W(A^2 - X^2) = 34- 30(.137*.137) = 33.4
b) ( X=0.137)
rate of work done = K*X*V = 450*(0.137)*33.4 = 2059.1 Watt
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