# A confusing energy question

1. Sep 30, 2007

### ZAMM

A child of mass 26kg is travelling at 4.0ms at the bottom of the slide which is 2m vertically below the top. The length of the slide is 6m.
Calculate

i) The energy used in overcoming the friction during the child's descent

ii) The magnitude of the frictional force.

Totally confused about this question. I know Work done = Force.displacement in the direction of the force but then does this mean that question i) is equal to W=26(9.8)Cos19.47?

I have no idea. Would appreciate any help.

thank you

2. Sep 30, 2007

### Staff: Mentor

Assume the child started at the top with zero speed. What's the total energy at the top? At the bottom?

3. Sep 30, 2007

### ZAMM

At the top would be the potential energy so 26*9.8*2, but this would be the energy output (I think). That's not really what i) is asking. At the bottom would also be the same but in kinetic energy form.

4. Sep 30, 2007

### Staff: Mentor

Why don't you check and see? They give you the speed. (The energy would only be the same if there were no losses to friction.)

5. Sep 30, 2007

### ZAMM

OK, so at the bottom of the slide E=.5(26)(4)^2 but I'm still confused. The length of the slide has 2 components? is the horizontal component the frictional component? I have no clue.

6. Sep 30, 2007

### turdferguson

Potential energy only deals with vertical displacement. You need the length of the slide for the second part of the question

7. Sep 30, 2007

### ZAMM

I know it only deals with vertical displacement but in this context the energy that requires him to get from the top to the bottom (neglecting the slide) would be mgh, but this alone does not help me solve the equation.

8. Sep 30, 2007

### Staff: Mentor

Good. How does that compare to the energy it started with at the top?

9. Sep 30, 2007

### turdferguson

The problem deals with conservation of energy. Etop=Ebottom. Without friction, PEtop=KEbottom, but thats not what you have here.

10. Sep 30, 2007

### ZAMM

At the top PE=mgh=26*9.8*2=509.6J and at the bottom KE=.5*26*4^2=208J. Energy is greater at the top than at the bottom.

11. Sep 30, 2007

### Staff: Mentor

OK, so what's the answer to the first question?

12. Sep 30, 2007

### ZAMM

The difference between them? but if it is that, what I calculated is not energy "used" but more wasted.

13. Sep 30, 2007

### Staff: Mentor

That depends on your point of view: You "used" it to overcome friction.