MHB Force of Friction: 0.5kg Ball & 0.381m Net

AI Thread Summary
The discussion revolves around calculating the force of friction between a net and a ball using the work-kinetic energy theorem. The ball exits the net at 0.600 m/s after entering at 4.5 m/s, and the net contacts the ball over a distance of 0.381 meters. Participants emphasize the need to incorporate both kinetic and potential energy changes in the calculations. Initial estimates for the force of friction vary, with some suggesting values around 9 N, while later calculations indicate that the correct answer might be closer to 17 N or even higher, depending on the initial speed used. The conversation highlights the importance of accurate data and unit consistency in physics problems.
Mango12
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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!
 
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Mango12 said:
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)
 
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 - - -

I like Serena said:
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 :/
 
Mango12 said:
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.
 
I like Serena said:
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 said:
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
 
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.
 
Mango12 said:
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
 
topsquark said:
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
 
  • #10
MarkFL said:
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
 
  • #11
topsquark said:
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.
 
  • #12
Mango12 said:
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?
 
  • #13
I like Serena said:
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
 
  • #14
I like Serena said:
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 - - -

I like Serena said:
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 - - -

MarkFL said:
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
 
  • #15
I like Serena said:
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 - - -

MarkFL said:
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
 
  • #16
Mango12 said:
...My teacher made a mistake. The ball actually enters the hoop at 6.42m/s and the answer should be 17N

Now, I get $F\approx31.7\text{ N}$.
 
  • #17
MarkFL said:
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?
 
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