# Power

• Oct 19th 2009, 06:05 PM
WhoCares357
Power
Quote:

A 0.55 kg mass sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (with k = 475 N/m) whose other end is fixed. The mass has a kinetic energy of 15.0 J as it passes through its equilibrium position (the point at which the spring force is zero).
Part 1:
Quote:

At what rate is the spring doing work on the mass as the mass passes through its equilibrium position?
I know the answer to this is 0W, but I have no idea how to get it.

Part 2:
Quote:

At what rate is the spring doing work on the mass when the spring is compressed 0.126 m and the mass is moving away from the equilibrium position?
I don't know how to find the velocity of the object. I can find F with F=-kx, but I need velocity to complete the equation for Power=F*v.
• Oct 19th 2009, 10:13 PM
Quote:

Originally Posted by WhoCares357
Part 1:

I know the answer to this is 0W, but I have no idea how to get it.

Part 2:

I don't know how to find the velocity of the object. I can find F with F=-kx, but I need velocity to complete the equation for Power=F*v.

HI

For (1)

you are told that when the object is in equilibrium position , the force of the spring exerted on the mass is 0 which also suggests that the work done on the mass is 0 (W=fs) , then it follows that power is also 0

For (2) , the elastic potential energy equals the kinetic energy of the mass from the principle of conservation of energy .

$\frac{1}{2}kx^2=\frac{1}{2}mv^2$