A 0.30 kg ladle sliding on a horizontal frictionless surface is attached to one

Question

A 0.30 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (k = 460 N/m) whose other end is fixed. The ladle has a kinetic energy of 14 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 0.11 m and the ladle is moving away from the equilibrium position?

Explanation / Answer

a) At the mean position there is no acceleration which means there is no force and hence at zero rate the spring is doing work.

b) at 0.1 m compressed we have force in the spring = kx = 500*.1 = 50 N

Now we need to calculate the velocity of the body at that instant.

This can be calculated using energy conservation

at mean position we have only KE but at the compressed position we have KE and PE

so

10 = 0.5*500*.1^2 + 0.5*0.34*v^2

v = 6.642 m/s

Hence rate of doing work = F.v = 50*6.642 = 332.11 W

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