Forces/conservation of energy of slide/free fall

[bdh2991]

Homework Statement

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.

Homework Equations

F = ma

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

 

[voko]

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

 

[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.

 

[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?

 

[bdh2991]

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

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

I think?

 

[voko]

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

 

[bdh2991]

Ok sooooo...

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

 

[voko]

Where is the reaction?

 

[bdh2991]

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

 

[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?

 

[bdh2991]

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

i'm sure this is wrong though

 

[voko]

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

 

[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

 

[voko]

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

 

[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)

 

[voko]

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

 

[bdh2991]

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

 

[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.

 

[bdh2991]

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

 

[voko]

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

 

[lep11]

This has very little to do with masses of the children, it is all about acceleration.
So you know that the acceleration of the smaller child is a=g=9.81m/s^2.
And as you said, the only force acting tangential to the slide is mg*sin(theta) since we ignore the friction. According to Newton's second law ƩF=ma→ mg*sin(theta)=m*a, so we get the acceleration of the other child a=g*sin(theta). Which one is bigger g or g*sin(theta)?