A sled is being pulled accross horizontal, snow covered ground

[smaan]

Homework Statement

A 16-kg sled is being pulled along the horizontal snow-covered
ground by a horizontal force of 24 N. Starting from rest, the sled attains
a speed of 2.0 m/s in 8.0 m. Find the coefficient of kinetic friction
between the runners of the sled and the snow.

Homework Equations

The Attempt at a Solution

I have found the acceleration of the sled by using the v^2 = v₀+2ax. So, 2^2 = 2(a)(8), which gives me 0.25 m/s^2. I also calculated the normal force. Since it is a horizontal plane, it would be (9.81)(16) = 156.96 N.

Not to sure how to get the μF_k from here.

Thank you for your time

 

[Doc Al]

How might you find the friction force acting on the sled? (Hint: Newton's 2nd law.)

 

[PeterO]

Homework Statement

A 16-kg sled is being pulled along the horizontal snow-covered
ground by a horizontal force of 24 N. Starting from rest, the sled attains
a speed of 2.0 m/s in 8.0 m. Find the coefficient of kinetic friction
between the runners of the sled and the snow.

Homework Equations

The Attempt at a Solution

I have found the acceleration of the sled by using the v^2 = v₀+2ax. So, 2^2 = 2(a)(8), which gives me 0.25 m/s^2. I also calculated the normal force. Since it is a horizontal plane, it would be (9.81)(16) = 156.96 N.

Not to sure how to get the μF_k from here.

Thank you for your time

If that is indeed the acceleration [I don't doubt you - I have just not checked your answer] you should know the net force, and thus how large the friction that opposes the applied 24N force.
Once you have the size of the friction force, you should be able to calculate the coefficient of friction, since you have the normal force.

 

[smaan]

How might you find the friction force acting on the sled? (Hint: Newton's 2nd law.)

Would the frictional force be F=ma? So would F=(16kg)(0.25m/s^2)=4N?

So then, μk=F_k/F_N. So, since I found the normal force to be 156.96 N, the kinetic friction coefficient would be 4/156.96 = 0.022548?

 

[PeterO]

Would the frictional force be F=ma? So would F=(16kg)(0.25m/s^2)=4N?

So then, μk=F_k/F_N. So, since I found the normal force to be 156.96 N, the kinetic friction coefficient would be 4/156.96 = 0.022548?

Newtons second law, using the actual acceleration, gives you the net Force acting.
The applied Force plus the Friction Force [adding as vectors of course] give the net Force.

The answer you give here is not correct.

 

[Doc Al]

Would the frictional force be F=ma? So would F=(16kg)(0.25m/s^2)=4N?

No. When you use F = ma, remember that F is the net force. So it's better to write it as: ƩF = ma.

There are two horizontal forces acting: The pulling force and the friction force. Set up an equation and solve for the friction force.

So then, μk=F_k/F_N. So, since I found the normal force to be 156.96 N, the kinetic friction coefficient would be 4/156.96 = 0.022548?

Once you solve for the correct friction force, then this method will give you μ_k.