# HELP Is my answer correct? Is negative Friction possible?

1. Apr 9, 2013

### 13physicsdude

HELP!! Is my answer correct?? Is negative Friction possible?!?

1. The problem statement, all variables and given/known data

A dogsled team has four dogs that pull a person and a sled with a combined mass of 100 kg.

A) They start from rest and reach a speed of 45 km/h in 2.5s. What is the average force applied by each dog?

B) Suppose each dog can pull with a force of magnitude 150 N. What is the frictional force acting on the sled

3. The attempt at a solution

A)
a=Vf-Vi/t
a=45-0/2.5
a=45/2.5
a=18 m/s2

Fnet=ma
Fnet=(100)(18)
Fnet=1800N/4 dogs
Fnet=450 N

∴ Average force applied by each dog is 450 N

B)
T-Ff=ma
600-Ff=ma
600-Ff=100(18)
600-Ff=1800
Ff=600-1800
Ff=-1200 N

∴ the Frictional force acting on the sled is -1200 N

Is my answer correct for part A and B ?? Its for an assignment due tomorrow and I want to make sure its correct.

2. Apr 9, 2013

### rude man

A is good, B is not.

In B you used an acceleration provided by dogs pulling a total of 1800N. In B though the total dog force is just 600N. So you know that computation is out the window.

In part B there is no acceleration. The force of 600N is used just to keep the sled moving. So in the absence of acceleration, what two forces balance each other?

EDIT: come to think of it, part A assumes zero friction, which is kind of weird considering part B involves friction, but there it is. Either that or the wording should have been "what is the average force, less friction force, applied by each dog?".

Also, ap123 is right, part A needs doing over to get the units right.

Textbook physics for you!

Last edited: Apr 9, 2013
3. Apr 9, 2013

### ap123

In part (a) you'll have to convert the speed from km/h to m/s.
This will give you an answer which makes more sense for part (b)

4. Apr 9, 2013

### Arkavo

following you should get friction coefficient=0.1

5. Apr 9, 2013

### 13physicsdude

Is ap123 correct? should I convert it from km/h to m/s?

6. Apr 9, 2013

### 13physicsdude

ya never mind ap123 was correct, I changed the units which gave me 5 m/s2= acceleration, which gave me 125 N force per dog for part "A".... I then multiplied by 4 because theres 4 dogs and I got 500N and plugged that into part "B" in replace of 1800N.... which gave me the final answer for part "B" of Ff= 100 N , is that correct now?

7. Apr 9, 2013

### ap123

That looks right :)
In part (b) each dog exerts an extra 25N making 100N extra compared with part (a).
To get the same acceleration, this 100N must be opposed by an equal frictional force

8. Apr 9, 2013

### rude man

Yes, that's right.

9. Apr 9, 2013

### 13physicsdude

alrighty, thanks everyone!! :)