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Physics Force of Friction

Mango12

Member
Jan 17, 2016
51
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!
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,491
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

Member
Jan 17, 2016
51
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 :/
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,491
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

Member
Jan 17, 2016
51
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

Member
Jan 17, 2016
51
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
MHB Math Helper
Aug 30, 2012
1,264
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

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
MHB Math Helper
Aug 30, 2012
1,264

Mango12

Member
Jan 17, 2016
51
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.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,491
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

Member
Jan 17, 2016
51
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

Member
Jan 17, 2016
51
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

Member
Jan 17, 2016
51
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

Administrator
Staff member
Feb 24, 2012
13,775
...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

Member
Jan 17, 2016
51
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?