Force of Friction: 0.5kg Ball & 0.381m Net

[Mango12]

Suppose that the ball exits the net going straight down at a velocity of just 0.600 m/s. What is the force of friction between the net and the ball, provided that the net contacts the ball throughout a length of 0.381 meters? The mass of the ball is 0.5kg. (The approximate answer is 9.0N)

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!

 

[I like Serena]

Suppose that the ball exits the net going straight down at a velocity of just 0.600 m/s. What is the force of friction between the net and the ball, provided that the net contacts the ball throughout a length of 0.381 meters? The mass of the ball is 0.5kg. (The approximate answer is 9.0N)

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!

Hi Mango12! Welcome to MHB! (Smile)

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 $F_{friction}$? (Wondering)

 

[Mango12]

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 :/

- - - Updated - - -

Hi Mango12! Welcome to MHB! (Smile)

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 $F_{friction}$? (Wondering)

I know...

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 :/

 

[I like Serena]

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 :/

So 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.

 

[Mango12]

So 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.

4.97/.381=13

 

[Mango12]

4.97/.381=13

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 9

 

[MarkFL]

I 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.

 

[topsquark]

4.97/.381=13

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

 

[MarkFL]

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

Wouldn't it actually be:

4.97 J / 0.381 m = 13 N

? :p

 

[topsquark]

Wouldn't it actually be:

4.97 J / 0.381 m = 13 N

? :p

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

-Dan

 

[Mango12]

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

-Dan

Right, units are always important. But is all the math right? Because the answer is supposed to be close to 9N. 13 isn't terribly far off, but I'm never very confident when it comes to math.

 

[I like Serena]

Right, units are always important. But is all the math right? Because the answer is supposed to be close to 9N. 13 isn't terribly far off, but I'm never very confident when it comes to math.

As 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?

 

[Mango12]

As 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?

I don't know. I just emailed my teacher asking him about it

 

[Mango12]

Hi Mango12! Welcome to MHB! (Smile)

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 $F_{friction}$? (Wondering)

Okay..so my teacher made a mistake. The correct answer should be about -17N

- - - Updated - - -

As 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?

Okay..so my teacher made a mistake. The correct answer should be about -17N

- - - Updated - - -

I 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.

Okay..so my teacher made a mistake. The correct answer should be about -17N

 

[Mango12]

Hi Mango12! Welcome to MHB! (Smile)

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 $F_{friction}$? (Wondering)

My teacher made a mistake. The ball actually enters the hoop at 6.42m/s and the answer should be 17N

- - - Updated - - -

I 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.

My teacher made a mistake. The ball actually enters the hoop at 6.42m/s and the answer should be 17N

 

[MarkFL]

...My teacher made a mistake. The ball actually enters the hoop at 6.42m/s and the answer should be 17N

Now, I get $F\approx31.7\text{ N}$.

 

[Mango12]

Now, I get $F\approx31.7\text{ N}$.

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?