- Thread starter
- #1

I know I have to use the work-KE theorem, and that w= the change in KE, but I really have no idea what to do. If it helps, the ball enters the hoop at 4.5m/s.

Thank you!

- Thread starter Mango12
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- Thread starter
- #1

I know I have to use the work-KE theorem, and that w= the change in KE, but I really have no idea what to do. If it helps, the ball enters the hoop at 4.5m/s.

Thank you!

- Admin
- #2

- Mar 5, 2012

- 9,491

Hi Mango12! Welcome to MHB!

I know I have to use the work-KE theorem, and that w= the change in KE, but I really have no idea what to do. If it helps, the ball enters the hoop at 4.5m/s.

Thank you!

This is indeed about the work-KE theorem.

The kinetic energy before has to be equal to the work applied by friction plus the kinetic energy after.

That is:

$$KE_{before}=W_{friction} + KE_{after}$$

$$\frac 12 m v_{before}^2 =F_{friction} \cdot s_{friction}+ \frac 12 m v_{after}^2$$

Can you perhaps fill in the numbers and solve for $F_{friction}$?

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- #3

1/2mv^2after = .5*.5*.6^2=0.09

But that's the only thing I know :/

- - - Updated - - -

I know...Hi Mango12! Welcome to MHB!

This is indeed about the work-KE theorem.

The kinetic energy before has to be equal to the work applied by friction plus the kinetic energy after.

That is:

$$KE_{before}=W_{friction} + KE_{after}$$

$$\frac 12 m v_{before}^2 =F_{friction} \cdot s_{friction}+ \frac 12 m v_{after}^2$$

Can you perhaps fill in the numbers and solve for $F_{friction}$?

1/2mv^2before = .5*.5*4.5^2=5.06

1/2mv^2after = .5*.5*.6^2=0.09

But that's the only thing I know :/

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- #4

- Mar 5, 2012

- 9,491

So we have:1/2mv^2before = .5*.5*4.5^2=5.06

1/2mv^2after = .5*.5*.6^2=0.09

But that's the only thing I know :/

$$5.06 = F_{friction} \cdot 0.381 + 0.09$$

What will $F_{friction}$ be?

Note that the work by friction is given by the force of friction times the distance that the force is active.

Or put simply: work = force times distance.

- Thread starter
- #5

4.97/.381=13So we have:

$$5.06 = F_{friction} \cdot 0.381 + 0.09$$

What will $F_{friction}$ be?

Note that the work by friction is given by the force of friction times the distance that the force is active.

Or put simply: work = force times distance.

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- #6

So then what do I do now? Because I got close to 13 when I tried it on my own but the answer is supposed to be close to 94.97/.381=13

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- #7

- Aug 30, 2012

- 1,264

I'm going to interrupt the conversation for a moment.4.97/.381=13

You are doing Physics here. Use units! It's 4.97 N / 0.381 = 13 N.

-Dan

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- #9

Wouldn't it actually be:I'm going to interrupt the conversation for a moment.

You are doing Physics here. Use units! It's 4.97 N / 0.381 = 13 N.

-Dan

4.97 J / 0.381 m = 13 N

?

- Aug 30, 2012

- 1,264

Heh. Just the kind of confusion I was talking about!Wouldn't it actually be:

4.97 J / 0.381 m = 13 N

?

-Dan

- Thread starter
- #11

Heh. Just the kind of confusion I was talking about!

-Dan

Right, units are always important. But is all the math right? Because the answer

- Admin
- #12

- Mar 5, 2012

- 9,491

As Mark already mentioned, we should also account for gravity (potential energy).Right, units are always important. But is all the math right? Because the answerissupposed to be close to 9N. 13 isn't terribly far off, but I'm never very confident when it comes to math.

That is:

$$

KE_{before} + PE_{before} = W_{friction} + KE_{after} + PE_{after}

$$

where $PE_{before} - PE_{after} = mg\Delta h$ and $W_{friction} = F_{friction}\Delta h$.

Either way, the force of friction will not come out as $9\text{ N}$.

If that is supposed to be the answer, I think there is some mistake in the given data.

Perhaps speed and kinetic energy were mixed up or some such?

- Thread starter
- #13

I don't know. I just emailed my teacher asking him about itAs Mark already mentioned, we should also account for gravity (potential energy).

That is:

$$

KE_{before} + PE_{before} = W_{friction} + KE_{after} + PE_{after}

$$

where $PE_{before} - PE_{after} = mg\Delta h$ and $W_{friction} = F_{friction}\Delta h$.

Either way, the force of friction will not come out as $9\text{ N}$.

If that is supposed to be the answer, I think there is some mistake in the given data.

Perhaps speed and kinetic energy were mixed up or some such?

- Thread starter
- #14

Okay..so my teacher made a mistake. The correct answer should be about -17NHi Mango12! Welcome to MHB!

This is indeed about the work-KE theorem.

The kinetic energy before has to be equal to the work applied by friction plus the kinetic energy after.

That is:

$$KE_{before}=W_{friction} + KE_{after}$$

$$\frac 12 m v_{before}^2 =F_{friction} \cdot s_{friction}+ \frac 12 m v_{after}^2$$

Can you perhaps fill in the numbers and solve for $F_{friction}$?

- - - Updated - - -

Okay..so my teacher made a mistake. The correct answer should be about -17NAs Mark already mentioned, we should also account for gravity (potential energy).

That is:

$$

KE_{before} + PE_{before} = W_{friction} + KE_{after} + PE_{after}

$$

where $PE_{before} - PE_{after} = mg\Delta h$ and $W_{friction} = F_{friction}\Delta h$.

Either way, the force of friction will not come out as $9\text{ N}$.

If that is supposed to be the answer, I think there is some mistake in the given data.

Perhaps speed and kinetic energy were mixed up or some such?

- - - Updated - - -

Okay..so my teacher made a mistake. The correct answer should be about -17NI think we must also account for the change in gravitational potential energy, and when I do, I get a rounded answer that is twice what you give as the intended answer of 9.0 N.

- Thread starter
- #15

My teacher made a mistake. The ball actually enters the hoop at 6.42m/s and the answer should be 17NHi Mango12! Welcome to MHB!

This is indeed about the work-KE theorem.

The kinetic energy before has to be equal to the work applied by friction plus the kinetic energy after.

That is:

$$KE_{before}=W_{friction} + KE_{after}$$

$$\frac 12 m v_{before}^2 =F_{friction} \cdot s_{friction}+ \frac 12 m v_{after}^2$$

Can you perhaps fill in the numbers and solve for $F_{friction}$?

- - - Updated - - -

My teacher made a mistake. The ball actually enters the hoop at 6.42m/s and the answer should be 17NI think we must also account for the change in gravitational potential energy, and when I do, I get a rounded answer that is twice what you give as the intended answer of 9.0 N.

- Admin
- #16

Now, I get $F\approx31.7\text{ N}$....My teacher made a mistake. The ball actually enters the hoop at 6.42m/s and the answer should be 17N

- Thread starter
- #17

That's what I got. But I know one of you said you got about 18N when you used the original 4.5m/s. Maybe I should just go back to doing that?Now, I get $F\approx31.7\text{ N}$.