Question

The mass of a child is 35kg. The child is raised to point A which is 1.5m and then released. She swings downwards through the equilibrium position B at 0.80m. What is the loss in gravitational potential energy of the child between A and B. And assuming that air resistance is neglible, calculate the speed of the child as she passes through the equilibrium position B.

Answer 1

At the top of the swing the child has gravitational potential energy, PE. At the bottom, she has all kinetic energy KE. That KE comes from the PE. Hence
KE(bottom) = PE(top).

Hence to solve this problem
1. Find PE(top)
2. Set it equal to KE(bottom)
3. Use the formula for KE to find speed, v

Answer 2

Make sense? Stuck anywhere?

Answer 3

To find the loss in PE would i use PE = Mgh?

Answer 4

Yes, PE is calculated as PE = mgh.

Answer 5

And what's the formula for KE?

Answer 6

Ok. To calculate The PE = 35*9.8*0.70?

Answer 7

Why is h = 0.7 m?

Answer 8

Never mind, I see it now in the problem: it's the difference 1.5 m - 0.8 m

Answer 9

That,s correct. Am i on the right track?

Answer 10

Yes

Answer 11

Hence the PE = 240 Joules

Answer 12

The second part of the question is that Ek=0.5*mv?

Answer 13

\[ KE = \frac{1}{2}mv^2 \] yes

Answer 14

Hence if PE = KE and PE = 240 J, then
\[ 240 = \frac{m}{2}v^2 \]
or
\[ v^2 = 240 \frac{2}{35} = 4202 \]

Answer 15

*apologies,
\[ v^2 = 240\frac{2}{35} = 13.72 \]

Answer 16

Is 13.72 ms the final answer or would you have to \[\S\] this?

Answer 17

Absolutely it is not the final answer. You are asked to find the speed of the child, so you must take the square root.

Answer 18

You need to write down a value for v.

Answer 19

square root EK/\[half\]m?

Answer 20

You have a value now for \( v^2 \). We know that
\[ v^2 = 13.72 \]

So to find \( v \), just take the positive square root of both sides of that equation. Why go backwards in the algebra to KE?

Answer 21

Hence what is the value of v?

Answer 22

Hello James. In the value 240 2/35 what is the 2 for?

Answer 23

We know that the kinetic energy, KE is given by
\[ KE = \frac{1}{2}mv^2 \]
We know that the KE(bottom) = PE(top). PE(top) = 240 J, hence
\[ KE = 240 \]
That is
\[ \frac{1}{2}mv^2 = 240 \]
hence
\[ v^2 = \frac{2}{m}240 = \frac{2}{35}240 \]

Make sense?

Answer 24

Yes. One more question if the rope stays taut throughout. Can you explain why the work done by the tension in the rope is zero?

Answer 25

Wait a second. What's your final value for v?

Answer 26

3.70ms?

Answer 27

3.70 m/s, yes.

Now I want to show you a quicker way to calculate it.

Answer 28

Ok carry on!

Answer 29

We know the PE at the top is equal to the KE at the bottom.

As before, let
h = vertical height at top of swing
v = speed at bottom.

Then as
PE = KE
we have

\[ mgh = \frac{1}{2}mv^2 \]
So far, so good?

Answer 30

Yes

Answer 31

Now we want to solve for v.

Notice the mass m cancels from both sides and we have
\[ v^2 = 2gh \]
Hence
\[ v = \sqrt{2gh} \]

Answer 32

Make sense?

Answer 33

The point is you now have just one equation to calculate, not two.

In your problem h = 0.7 m, hence
\[ v = \sqrt{2gh} = \sqrt{2(9.8)(0.7)} = \sqrt{13.72} = 3.70 \]
End of problem.

Answer 34

This formula
\[ v = \sqrt{2gh} \] is very handy.

Answer 35

I know James i have done this equation before and got the same answer but i was'nt sure i was doing it correct though!

Answer 36

Well now you do.

Second part of your question. What is the equation for work done by a force F?

Answer 37

Yes that,s right!

Answer 38

I'm addressing now your question: "Can you explain why the work done by the tension in the rope is zero?"

So first, what is the definition of Work?

Answer 39

Work = Pt

Answer 40

Nope, that's not the definition. The definition is work is the product of Force acting over a distance d, where we only consider the Force in the direction of d.

So if I pick up a 1 kg book from the floor and put it on a table 1 m height, the force is
F = mg
and I acted against a distance of
d = 1 meter

hence the work I did was
Work = Fd = 1(9.8)(1) = 9.8 J

Answer 41

Now if I pick up the book at hold it one meter off the floor and walk around with it for 8 hours, keeping the book all the time at 1 meter. How much work have I done?

Answer 42

And I walk 20 kilometers ... how much work have I done on the book?

Answer 43

still there?

Answer 44

If you don't know, at least guess. And be quick. Other people are asking for me.

Answer 45

Sorry james got distracted thanks for your help.