Kinetic Energy and Newton Forces - distance/speed

[Lori]

Homework Statement

A sled of mass m is given a kick on a frozen pond. The kick imparts to it an initial speed of 2.20 m/s. The coefficient of kinetic friction between sled and ice is 0.150. Use energy considerations to find the distance the sled moves before it stops

Homework Equations

Friction Force = Friction coefficient * Normal Force
Normal Force = mass * g
W = KE = 1/2mv^2
W=fd

The Attempt at a Solution

So, I understand that force of friction is equal to mg*Friction coefficient (given that N=mg)

But, why is mgu=.5mv^2 (final Kinetic )

Shouldn't it be that the work is work done by applied force minus work done by Friction?

Why only look at friction work?

 

[Doc Al]

Shouldn't it be that the work is work done by applied force minus work done by Friction?

The work done by the applied force -- the kick -- is what gave the sled its initial velocity. It doesn't act beyond that. (The kick is just for an instant.)

 

[Lori]

The work done by the applied force -- the kick -- is what gave the sled its initial velocity. It doesn't act beyond that. (The kick is just for an instant.)

Oh ok, so all the work net is coming from the friction force which is equal to coefficient friction * mass *gravity * distance. And, since work net is equal to total kinetic energy which is just .5mv^2, we set those equal to each other.

 

[Doc Al]

Oh ok, so all the work net is coming from the friction force which is equal to coefficient friction * mass *gravity * distance. And, since work net is equal to total kinetic energy which is just .5mv^2, we set those equal to each other.

Right. Another way to word it: The net work done equals the change in KE. (Both the work done by friction and the change in KE are negative.)