# Work and Work-Energy Theorum

1. Jul 11, 2010

### mizzy

1. The problem statement, all variables and given/known data
A 6kg object is lifted vertically through a distance of 5m by a light string under a tension of 80.0N. Find (a) The work done by the force of tension, (b) The work done by gravity, and (c) The final speed of the object if it starts from rest.

2. Relevant equations
W = Fd

Work net = change in kinetic energy

3. The attempt at a solution
I think I got part a and b.

For part a,
W = Fd
where, F is the tension given and distance is 5m.

For part b,
W = Fd
where, F is equal to mg times 5m.

For part c,
I'm not sure if I did this one right. The work-energy theorum says that the net work done on an object is equal to the change in kinetic energy. So for net work I took the work done by the force of tension(part a) minus the work done by gravity(part b) is equal to the change in kinetic energy. I then solved for vfinal.

Did i do part c correctly?

2. Jul 11, 2010

### vela

Staff Emeritus
Your answer to part (b) should be negative since the displacement is in the direction opposite the force. For a constant force F, W=Fd only if the force and displacement are in the same direction; more generally, W=Fd cos θ, where θ is the angle between the force and the displacement.

In part (c), you would then just add the two contributions together. Since one is positive and one is negative, they partially cancel each other out. The leftover is the change in kinetic energy.

3. Jul 11, 2010

### mizzy

Thanks for telling me about part b. I forgot that work will be negative if the force is opposite to the displacement.

Ok. I think I got it now.