A 0.55 kg mass sliding on a horizontal frictionless surface is attached to one e
ID: 1331439 • Letter: A
Question
A 0.55 kg mass sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (with k = 600 N/m) whose other end is fixed. The mass has a kinetic energy of 13.0 J as it passes through its equilibrium position (the point at which the spring force is zero).
a) At what rate is the spring doing work on the mass as the mass passes through its equilibrium position?
b) At what rate is the spring doing work on the mass when the spring is compressed 0.104 m and the mass is moving away from the equilibrium position?
Explanation / Answer
a)
0 W
At equilibrium, there is no force from the spring acting on the mass. As work is Force times distance, with no force, no work is done and no power is applied as power is the rate of doing work.
b)
The force on the spring at this compression is
F = kx
F = 600 * 0.104
F = 62.4 N
Use energy equations to find the velocity
13 = ½kx² + ½mv²
½mv² = 13 - ½kx²
v = ((2(13 - ½kx²)) / m)
so when compression is 0.104 m
v = ((2(13 - ½600(0.104²)) / 0.55)
v = 5.96 m/s
velocity is distance over time
v = d/t
work is force times distance
W = Fd
Power = Work / time
P = Fd/t
P = 62.4 * (5.96)
Rate of work = Power = 372 Watts
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