Work Example

A box with a mass of 1.0 kg is moving to the right. Its initial speed is 2 m/s. After traveling 3 m, its speed is 4 m/s. In addition to the usual forces (gravity, normal, friction) there are two other forces acting on the box. One is a 5 N force that has an upward vertical component of 4 N and a component to the left of 3 N. The other force is 8 N to the right.

(a) What is the normal force?

(b) What is the coefficient of kinetic friction?

A good free-body diagram is needed here.


Applying Newton's second law in the y-direction, it can be shown that the normal force is
N = 9.8 - 4 = 5.8 N.

Applying Newton's second law in the x-direction tells us:

ΣFx = 8 - 3 - fk = 5 - fk = max

Finding μk takes more work. Let's try the energy method.

Energy method

Start with the master energy equation:

Ui + Ki + Wnc = Uf + Kf

There is no change in potential energy, so we can get rid of the potential energy terms. This gives:

Ki + Wnc = Kf

Wnc = Kf - Ki = ½ mvf2 - ½ mvo2

Wnc = 8 J - 2 J = 6 J

In this case Wnc = 6 = (ΣFx) x (3 m), so:
ΣFx =
6
3
= 2 N

We showed already that ΣFx = 5 - fk, so:

fk = 5 - ΣFx = 5 - 2 = 3 N

We found the normal force to be 5.8 N, so:
μk =
fk
N
=
3
5.8
= 0.52

Could we have done this problem using constant acceleration equations? Absolutely. We would have found the acceleration from the equation:

vx 2 = vox2 + 2 ax (x - xo)

If we multiply through by ½ m, the equation becomes:

½ mvx 2 = ½ mvox2 + max (x - xo)

This says Kf = Ki + Fx Δx, which is our master energy equation without the potential energy terms.

That constant acceleration equation is actually an energy equation in disguise - so we've been using energy concepts all along.