How much work is done by frictional force on a block moving toward equilibrium?

[c4iscool]

Homework Statement

A 20-kg block on a horizontal surface is attached to a light spring (force constant = 8.0 kN/m). The block is pulled 10 cm to the right from its equilibrium position and released from rest. When the block has moved 2.0 cm toward its equilibrium position, its kinetic energy is 13 J. How much work is done by the frictional force on the block as it moves the 2.0 cm?

do I need a coefficient of friction to solve this?

 

[kreil]

You are told what the blocks kinetic energy is with friction, so if you solve for what it should be without friction then the difference is the work done by the frictional force...no coefficient of friction needed.

 

[c4iscool]

ok, I get that but how would I find the velocity to get the kinetic energy?

ke = 1/2 M*V^2

or am I still not getting it?

 

[PhanthomJay]

ok, I get that but how would I find the velocity to get the kinetic energy?

ke = 1/2 M*V^2

or am I still not getting it?

You need to apply the total conservation of energy principle. Are you familiar with it?

 

[c4iscool]

ummm...no. I'm about to google it tho, but please explain.

 

[PhanthomJay]

ummm...no. I'm about to google it tho, but please explain.

In the absence of non conservative forces other than friction, you should discover that $$\Delta K + \Delta U + W_{friction}= 0$$. Now you must apply it.

 

[c4iscool]

I got it. I think it works out to be -1.4J

 

[PhanthomJay]

I got it. I think it works out to be -1.4J

Looks right. Even your minus sign is correct. I slipped on mine, sorry, I should have said $$\Delta K + \Delta U - W_{friction} = 0$$