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:

ΣF_x = 8 - 3 - f_k = 5 - f_k = ma_x

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

Energy method

Start with the master energy equation:

U_i + K_i + W_nc = U_f + K_f

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

K_i + W_nc = K_f

W_nc = K_f - K_i = ½ mv_f² - ½ mv_o²

W_nc = 8 J - 2 J = 6 J

In this case W_nc = 6 = (ΣF_x) x (3 m), so:

ΣFx

=

6 3

=

2 N

We showed already that ΣF_x = 5 - f_k, so:

f_k = 5 - ΣF_x = 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:

v_x ² = v_ox² + 2 a_x (x - x_o)

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

½ mv_x ² = ½ mv_ox² + ma_x (x - x_o)

This says K_f = K_i + F_x 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.