# PhysicsForce of Friction

#### Mango12

##### Member
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
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!

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 $F_{friction}$?

#### Mango12

##### Member
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!

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 $F_{friction}$?
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
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
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
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

Staff member
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
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

Staff member
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

?

#### topsquark

##### Well-known member
MHB Math Helper
Wouldn't it actually be:

4.97 J / 0.381 m = 13 N

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

-Dan

#### Mango12

##### Member
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 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
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
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
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 $F_{friction}$?

- - - 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?

- - - 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.

#### Mango12

##### Member
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 $F_{friction}$?
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

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