Find coefficient of kinetic friction using energy conservation

[offbeatjumi]

Homework Statement

An object of mass 0.2 kg with initial velocity 4.8 m/s slides a horizontal distance of 3.0 m before coming to rest (final velocity is 0 m/s). What is the coefficient of kinetic friction on the surface?

Homework Equations

K + U - W(other) = 0 ...conservation of energy

The Attempt at a Solution

I have delta-K = 0 - (1/2)(0.2)(4.8)^2 = -0.48 J
delta-U is zero since there is no change in y
W(other) would be just the work done by the frictional force f(k).
f(k) <= (mu)kn...where n-mg = 0... so n = mg = 1.96 N ...f(k) <= (mu)k*1.96N

W = F*d... F = f(s)...
since U = 0 ... K - W(other) = 0... K = W(other) = -0.48 J
F*d = -0.48 J
(mu)k*1.96N*3.0m = -0.48 J

Solving for (mu)k I get (mu)k = 0.0816.

From the back of the book I know this is wrong, the given answer is 0.392.
What am I doing wrong?
Thanks!

 

[Bacat]

The first thing that jumps out at me is this:

\frac{1}{2}*0.2*4.8^2 = -0.48 J

By my calculation, that is not true.

 

[jamesrc]

Try recalculating the kinetic energy - the number you got is wrong (you didn't square the velocity).

Also, when you say that the friction is less than or equal to mu_k*N - it's really just equal (that is, f = mu_k*N). The less-than-or-equal to part comes into play when you are considering static friction, where f<=mu_s*N (so that as long as the applied force does not exceed the maximum static friction, the object will not move and the friction force will be equal in magnitude to the applied force)

 

[offbeatjumi]

thank you so much, that was a stupid mistake.