Why why why is Potential Energy equal to Kinetic Energy in this problem?

1. Oct 23, 2012

riseofphoenix

Why why why is Potential Energy equal to Kinetic Energy in this problem??

A 31.0 kg child on a 3.00 m long swing is released from rest?
when the ropes of the the swing make an angle of 28.0° with the vertical

(a) Neglecting friction, find the child's speed at the lowest position.____m/s

potential energy = kinetic energy
mgh = 1/2 mv2
0.35 * 9.8 = 0.5 v2
v = √(2*9.8*0.35)
v = 2.62 m/s (without friction)

^^^^^^^

Why did they set potential energy equal to kinetic energy? I don't understand :(

2. Oct 23, 2012

Staff: Mentor

Re: Why why why is Potential Energy equal to Kinetic Energy in this problem??

Mechanical energy is conserved:

KEA + PEA = KEB + PEB

Let A be the highest point, where the swing is released. Thus, KEA = 0.

If you measure PE from the lowest point (point B), then PEB = 0.

That gives you:

PEA = KEB

3. Oct 23, 2012

riseofphoenix

Re: Why why why is Potential Energy equal to Kinetic Energy in this problem??

What do you mean by "mechanical energy" though?

4. Oct 23, 2012

Staff: Mentor

Re: Why why why is Potential Energy equal to Kinetic Energy in this problem??

Mechanical energy means the sum of KE + PE.

I assume you've been studying conservation of energy?

5. Oct 23, 2012

riseofphoenix

Re: Why why why is Potential Energy equal to Kinetic Energy in this problem??

Yes but I'm a little confused...ok so...

The girl is swinging.

When the swing is at an angle of 28 degrees, (slow mo) she stops, and at that point you have potential energy....

When the swing moves forward again you no longer have potential energy (PE = 0), but this time, you have Kinetic energy...

So, essentially, this problem deals with the law of conservation of energy, right? which is:

PEA + [STRIKE]KEA [/STRIKE]= [STRIKE]PEB[/STRIKE] + KEB

So that's why, PEA = KEB, right?
Which is,

mgh = (1/2)mv2

Right?

If so, then what do I did with the angle (theta) that they gave me? Where does that go in the equation above?

Last edited: Oct 23, 2012
6. Oct 23, 2012

Staff: Mentor

Re: Why why why is Potential Energy equal to Kinetic Energy in this problem??

Right.

No, not right. The potential energy here is gravitational PE (mgh) not spring PE (which is 1/2kx2). No springs in this problem!
Once you have the correct expression for PE, you'll need the angle to figure out the height of the initial position.

7. Oct 23, 2012

riseofphoenix

Re: Why why why is Potential Energy equal to Kinetic Energy in this problem??

mgh(1 - cos 28) = (1/2)mv2

I looked that up :(
Now my question is, would it ALWAYS be h(1 - cos θ) whenever I have a problem like this?

8. Oct 23, 2012

riseofphoenix

Re: Why why why is Potential Energy equal to Kinetic Energy in this problem??

Sorry if I'm asking so many questions - I'm just trying to figure out how these things relate to each other so that I can go about answering any question like this!

Part b says:

If the speed of the child at the lowest position is 2.30 m/s, what is the mechanical energy lost due to friction?

What they did:

KE @ 2.62 m/sec = (1/2)(31)(2.62)^2 = 106.68 J

KE @ 2.30 m/sec = (1/2)(31)(2.30)^2 = 81.995 J

Therefore, energy lost to friction = 106.68 - 81.995 = 24.685 J

^^^^^^

My question is, how did they know to just subtract KE with v=2.30 by KE with 2.62?

9. Oct 24, 2012

Staff: Mentor

Re: Why why why is Potential Energy equal to Kinetic Energy in this problem??

That depends on just how 'like' the problem is, doesn't it? Get the concept: To find the change in gravitational PE, you may need to determine the change in height. By whatever means necessary.

10. Oct 24, 2012

Staff: Mentor

Re: Why why why is Potential Energy equal to Kinetic Energy in this problem??

They are really subtracting final energy from initial energy:

Initial Energy (at top) = Final Energy (at bottom) + Energy lost to friction

11. Oct 24, 2012

CWatters

Re: Why why why is Potential Energy equal to Kinetic Energy in this problem??

No, that comes from the geometry of the problem.

In general.. ΔPE = mgΔh

So you need to work out the change in height (Δh) using whatever information is given in the problem. You might have a similar problem where θ is specified differently. If in doubt make your own drawing.