# Homework Help: Work by Gravity

1. Oct 24, 2007

### BuBbLeS01

1. The problem statement, all variables and given/known data
A 0.113-kg ball is thrown straight up from 2.07 m above the ground. It's initial vertical speed is 7.6 m/s. A short time later, it hits the ground. Calculate the total work done by the force of gravity during that time.

2. Relevant equations

3. The attempt at a solution

can I use...
1/2mv^2 = mgh
1/2v^2 - gh = W

2. Oct 24, 2007

### G01

Think of the problem like this:

What is the initial energy of the ball? (When it is thrown)

What is the final energy of the ball? (When it lands)

If you know those two quantities, you should be able to find the work.

3. Oct 24, 2007

### Vijay Bhatnagar

Work done = Force x Displacement. What is gravitational force acting on the ball? What is the displacement (not distance travelled) between the initial position and final positions of the ball?

4. Oct 25, 2007

### BuBbLeS01

W=Fd
gravitational force = g
displacement = 2.07m

5. Oct 31, 2007

### BuBbLeS01

Okay so I am not understanding this problem. Apparently from 2.07m and above there is no work done but I don't know why?? And I get...
W = Fd
Ug = mgh
K = 1/2mv^2
Wtot = mgh = 2.29J

I was trying to do W = Change in K + change in Ug....why is this not the case?

6. Oct 31, 2007

### hotcommodity

The total work done on the ball is given by the ball's overall change in kinetic energy. When the ball goes up from its initial height, and reaches the top of its path, its kinetic energy has decreased. When it comes down from the top and back to that same initial height, its kinetic energy has increased. The decrease and increase in kinetic energy will add to zero, because we're ignoring air resistance. You want to find the change in the ball's kineteic energy from 2.07m to the ground. Does that help?

7. Oct 31, 2007

### BuBbLeS01

Oh okay...thank you that makes sense!!