# Homework Help: Work, Energy, and Power problems

1. Jan 17, 2008

### Twilit_Truth

Ok, this is the last day I have to work on this stuff, so please bear with me. I'll be posting questions in this thread as needed, so check back in every once in a while please.

A pole vaulter clears 6.00 m. With what velocity does the vaulter strike the mat in the landing area? (Disregard air resistance. g = 9.81 m/s^2)

2. Jan 17, 2008

### rock.freak667

At 6.00m what is the vaulter's gravitational p.e.? Since you are disregarding air resistance, what is all the energy going to be converted to?

3. Jan 17, 2008

### Twilit_Truth

But PE = mgh, and I don't have m.

4. Jan 17, 2008

### rock.freak667

While that is true, as the vaulter falls, kinetic energy is gained while potential energy is lost. What does that mean now?

5. Jan 17, 2008

### Twilit_Truth

mgh = 1/2 mv^2. Masses cancel. Thanks.

6. Jan 17, 2008

### Twilit_Truth

New problem.

A 40.0 N crate starting at rest slides down a rough 6.0 m long ramp inclined at 30.0 degrees with the horizontal. The force of friction between the crate and the ramp is 6.0 N. Using the work-kinetic energy theorem, find the velocity of the crate at the bottom of the incline.

7. Jan 17, 2008

### rock.freak667

Well at the top of the 6m incline the crate has PE, and as it slides down all the PE is converted to KE and work done in overcoming friction.

8. Jan 17, 2008

### Twilit_Truth

work done in overcoming friction?

9. Jan 17, 2008

### rock.freak667

Yes, if there is not sufficient energy to overcome friction, then the block won't move. But the work done in overcoming friction is just simply Force*Distance

10. Jan 17, 2008

### Twilit_Truth

And force is the downward force and distance is 6.0 m?

11. Jan 17, 2008

### rock.freak667

The frictional force is the opposite direction of the motion of the crate.So it is the frictional force*distance(6)

12. Jan 17, 2008

### Twilit_Truth

The frictional force you speak of is 6.0 N, right?

13. Jan 17, 2008

### rock.freak667

Indeed

14. Jan 17, 2008

### Twilit_Truth

Ok, I did this:

mgh=PE
PE=(40.0)(9.81)(3)=1177.2 J

I got the height by this:

6.0 * sin(30)=3

then:

PE=KE+W
PE-W=KE
W=F(6.0 N)*d(6.0 m)=36 J
1177.2-36=1141.2 J

That might be where I messed up, but I kept going:

KE=1/2mv^2
2KE=mv^2
2KE/m=v^2
(2(1141.2))/40.0=v^2
57.06=v^2

So, the square root of 57.06 equals velocity, but it doesn't match the answers I have. It is a multiple choice question, here are the anwers:

A) 8.7 m/s

B) 3.3 m/s

C) 4.5 m/s

D) 6.4 m/s

What did I do wrong?

15. Jan 17, 2008

### rl.bhat

Here m is not equual to 40. mg = 40N. SO m = 40/9.81 kg