Why is potential energy equal to kinetic energy for a swinging child?

[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 :(

 

[Doc Al]

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

Mechanical energy is conserved:

KE_A + PE_A = KE_B + PE_B

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

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

That gives you:

PE_A = KE_B

 

[riseofphoenix]

Mechanical energy is conserved:

KE_A + PE_A = KE_B + PE_B

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

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

That gives you:

PE_A = KE_B

What do you mean by "mechanical energy" though?

 

[Doc Al]

What do you mean by "mechanical energy" though?

Mechanical energy means the sum of KE + PE.

I assume you've been studying conservation of energy?

 

[riseofphoenix]

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:

PE_A + [STRIKE]KE_A [/STRIKE]= [STRIKE]PE_B[/STRIKE] + KE_B

So that's why, PE_A = KE_B, 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?

 

[Doc Al]

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

PE_A + [STRIKE]KE_A [/STRIKE]= [STRIKE]PE_B[/STRIKE] + KE_B

So that's why, PE_A = KE_B, right?

Right.

Which is,

(1/2)kx² = (1/2)mv²

Right?

No, not right. The potential energy here is gravitational PE (mgh) not spring PE (which is 1/2kx²). No springs in this problem!

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

Once you have the correct expression for PE, you'll need the angle to figure out the height of the initial position.

 

[riseofphoenix]

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?

 

[riseofphoenix]

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?

 

[Doc Al]

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?

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.

 

[Doc Al]

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

They are really subtracting final energy from initial energy:

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

 

[CWatters]

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?

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.