# Homework Help: Simple energy/work

1. Jan 27, 2010

### tigerwoods99

1. The problem statement, all variables and given/known data
A 890 N crate rests on the floor.
How much work is required to move it at constant speed 5.6 m along the floor against a friction force of 180 N?
How much work is required to move it at constant speed 5.6 m vertically?

2. Relevant equations
KE = (1/2)mv^2

3. The attempt at a solution
(1/2)(890/9.8) = x?

2. Jan 27, 2010

### Phyisab****

You don't know the velocity so kinetic energy won't help you. Any other equations you can think of?

3. Jan 27, 2010

### tigerwoods99

Where:
Fa = applied force
D = displacement
Cos(theta= im not sure would it be pie/2

4. Jan 27, 2010

### Phyisab****

cos(pi/2) = 0, here your work is being done in the same direction as the motion so.

That equation gets you half way there, need one more equation.

5. Jan 27, 2010

### tigerwoods99

I think cos(pi/2) = 1.?
Mhm is the other equation:
Work done by friction = FfD
Where Ff= friction force
D = displacement?

6. Jan 27, 2010

### Phyisab****

Actually I wasn't quite right all you need is the equation you just wrote sorry bout that. So just use the equation you just wrote, (W done by friction) = (F of friction)*d
Then just plug and chug.

7. Jan 27, 2010

### tigerwoods99

Thanks I was able to answer part 1.
How much work is required to move it at constant speed 5.6 m vertically?

So here there is a gravitational component?

8. Jan 27, 2010

### Phyisab****

W=f*d, you know the distance. Can you calculate the force due to gravity?

9. Jan 27, 2010

### tigerwoods99

The force due to gravity is just mg, (860)
W = 890*5.6?

10. Jan 27, 2010

### Phyisab****

there ya go easy right? In general if you have an equation with n variables you need n-1 equations to solve for the one variable you're looking for.

11. Jan 27, 2010

### tigerwoods99

Thanks, and thanks for the tip!