Question

A .35 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (with the spring constant (k) =455 N/m) whose other end is fixed. The mass has a kinetic energy of 10 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 ladle as the ladle passes through its equilibruim position?

And

At what rate is the spring doing work on the ladle when the spring is compressed .1m and the ladle is moving away from the equilibrium position?

Answer 1

I presume you are in calculus based physics.

Rate of work= d work/dt= d/dt (1/2 kx^2)= kx dx/dt = kx * velocity

But velocity= sqrt(2*KE/mass)

where KE= 10-1/2 kx^2 so

<b>rate of work= kx*(sqrt2/mass (10-1/2kx^2)</b>

check my thinking.

Answer 2

ok..that helps..thanks