HELP Is my answer correct? Is negative Friction possible?

[13physicsdude]

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

Homework Statement

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

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 NB)
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.

 

[rude man]

Homework Statement

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

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.

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!

 

[ap123]

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

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)

 

[Arkavo]

following you should get friction coefficient=0.1

 

[13physicsdude]

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

 

[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 there's 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?

 

[ap123]

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 there's 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?

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

 

[rude man]

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 there's 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?

Yes, that's right.

 

[13physicsdude]

alrighty, thanks everyone! :)