# Forces/conservation of energy of slide/free fall

1. Oct 18, 2012

### bdh2991

1. The problem statement, all variables and given/known data

Two children stand on a platform at the top of a curving
slide next to a backyard swimming pool. At the same
moment the smaller child hops off to jump straight down
into the pool, the bigger child releases herself at the top
of the frictionless slide. During their motions from the platform
to the water, the average acceleration of the smaller child
compared with that of the larger child is (a) greater (b) less
(c) equal.

2. Relevant equations
F = ma

3. The attempt at a solution

I originally though it was equal because if you use F= ma, the only force acting on them is gravity right? so it it would be mg = ma, therefore a = g, but i don't understand how the answer could be a. could someone help me with this and represent it mathematically...thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 18, 2012

### voko

Consider the change in the potential energy and the length of the path traveled by either child.

3. Oct 18, 2012

### bdh2991

Ok so initially both of them have energy of mgh, and Mgh (for the bigger one)...but they both travel the same height, so the change in potential energy would be larger for the child with more mass.

4. Oct 18, 2012

### voko

They travel the same height, but not the same path. But when I think about this approach again, it seems too complex. Let's try something else.

Imagine a child somewhere on the slide. What are the forces acting on him, and what is his acceleration?

5. Oct 18, 2012

### bdh2991

Y-Forces would be N*cos(theta) - mg = ma

X-Forces would be N*sin(theta) = ma

I think?

6. Oct 18, 2012

### voko

No, that's not right. Use the coordinates normal and tangential to the slope.

7. Oct 18, 2012

### bdh2991

Ok sooooo....

Y-forces would be F = mg*cos(theta) = ma
X-forces would be F = mg*sin(theta) = ma

8. Oct 18, 2012

### voko

Where is the reaction?

9. Oct 18, 2012

### bdh2991

the reaction of the force would along the horizontal
so F = sqrt((mgcos(theta)^2 + mgsin(theta)2)) = 2mg?

10. Oct 18, 2012

### voko

I do not follow.

Do a FBD. Write down the components normal and tangential to the slide. Do not forget the reaction. What is the acceleration?

11. Oct 18, 2012

### bdh2991

Ax = gsin(theta)
Ay = gcos(theta)

i'm sure this is wrong though

12. Oct 18, 2012

### voko

Again, use normal/tangential axes. Acceleration can only be tangential.

13. Oct 18, 2012

### bdh2991

i don't get it honestly...acceleration has to have 2 components to it when it is on a slope if your asking for the resultant acceleration then i suppose that would be sqrt(Ax^2 + Ay^2) = g

14. Oct 18, 2012

### voko

Do you understand what "normal" and "tangential" mean?

15. Oct 18, 2012

### bdh2991

normal is perpendicular to the surface tangential is along the surface:

the acceleration is along the tangential so that would be the x-component of the force so it would be mg*sin(theta)

16. Oct 18, 2012

### voko

Correct. Now compare that with acceleration of the child plunging straight down.

17. Oct 18, 2012

### bdh2991

so mg*sin(theta) < mg....therefore because of a larger force the acceleration would also have to be larger?

18. Oct 18, 2012

### voko

Force and acceleration are very simply related, aren't they?

Note the formula above has a problem: it uses the same "m", while you have two different masses.

19. Oct 18, 2012

### bdh2991

ok so if we don't know the size of each m how could you even tell which is greater?

20. Oct 18, 2012

### voko

What you have is forces; you are asked to compare accelerations.