A 0.54 kg ladle sliding on a horizontal frictionless surface is attached to one
ID: 1319366 • Letter: A
Question
A 0.54 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (k = 550 N/m) whose other end is fixed. The ladle has a kinetic energy of 240 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 ladle as the ladle passes through its equilibrium position? (b) At what rate is the spring doing work on the ladle when the spring is compressed 0.12 m and the ladle is moving away from the equilibrium position?
Explanation / Answer
1.
Kinetic energy at equilibrium point =1/2*m*v^2
KE = 1/2 * 0.54 * v^2 = 240
v^2 = 888.888
v = 29.8 m/s
As the ladle is moving at this speed, the spring will come in to action.
However at the equilibrium position the spring tension = zero
Hence at the instant the ladle is passing the equilibrium position work done = zero
2.
At 0.12 m, tension force in the spring = 550 * 0.12 = 66 N
We assume the velocity remains the same (strictly velocity will decrease due to deceleration imposed by the spring)
Rate of Work done = 66 N * 29.8 m/s = 1966.8 Nm/s = 1966.8 W
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