Energy Watch

keevesygoesrawr
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The diagram shows a child's swing. A child of mass of 28kg is released from rest at position A. Assuming that there are no resistive forces, calculate the

(I) kinetic energy of the child when at point B

(ii) speed of the child at B

The diagram has an angle of 45 degrees and is labelled 1.6m on both A and B.

I know the equation for KE but that's all.. please help!

1/2 mv(2)
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Stonebridge
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(Original post by keevesygoesrawr)
The diagram shows a child's swing. A child of mass of 28kg is released from rest at position A. Assuming that there are no resistive forces, calculate the

(I) kinetic energy of the child when at point B

(ii) speed of the child at B

The diagram has an angle of 45 degrees and is labelled 1.6m on both A and B.

I know the equation for KE but that's all.. please help!

1/2 mv(2)
You either need to post the diagram. (preferable)
or
explain more fully
- where point B is
- what the angle 45 degrees measures
- what 1.6m measures?

We could guess but that won't help you very much.
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keevesygoesrawr
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(Original post by Stonebridge)
You either need to post the diagram. (preferable)
or
explain more fully
- where point B is
- what the angle 45 degrees measures
- what 1.6m measures?

We could guess but that won't help you very much.
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Stonebridge
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Find the vertical height of A above B.
Spoiler:
Show

Construct a right angled triangle by drawing a horizontal line from A across to the vertical line.
Use Pythogaras to calculate what you need.


Then use loss in potential energy falling from A to B through that vertical height equals gain in kinetic energy.
As k.e. was zero at A it gives you the k.e and hence speed at B.
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ohnanailikenanas
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@stonebridge Could you explain it to me? Or could you tell me the answer so I can work it out from there?


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harrysingh
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someone give me june 2014 unit 4 mark scheme
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ohnanailikenanas
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For what Exam board..


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Stonebridge
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(Original post by harrysingh)
someone give me june 2014 unit 4 mark scheme
This is not allowed, unfortunately.
Exam papers and mark schemes are not released until a year after the examination finished.
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Stonebridge
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(Original post by ohnanailikenanas)
@stonebridge Could you explain it to me? Or could you tell me the answer so I can work it out from there?


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What don't you understand from what I posted above?
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ohnanailikenanas
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1.6(2) + 1.6(2) = 6.4
6.4 Cos(45) = 3.4m
Now what do I do?
KE - 1/2mv(2)
1/2 x 28 x v(2)

Dunno how to find the speed or KE


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Stonebridge
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(Original post by ohnanailikenanas)
1.6(2) + 1.6(2) = 6.4
6.4 Cos(45) = 3.4m




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What is this?
I can't figure out where you get
1.6(2) + 1.6(2) from.

You need to find the vertical height of the swing at A above the position at B.
It can't possibly be 3.4m
The swing is only 1.6m long.
I put the method in the spoiler.
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ohnanailikenanas
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I tried Cos(45) x 1.6 = 72
I'm not sure :/


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harrysingh
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aqa fam
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Stonebridge
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(Original post by ohnanailikenanas)
I tried Cos(45) x 1.6 = 72
I'm not sure :/


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Cos (45) = 0.707
so how can cos (45) x 1.6 = 72 ????

Subtract the correct answer for this from 1.6 to get the difference between the heights of A and B
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