Select the correct answer.

A force of 16.0 newtons is sufficient to set a box that is at rest into motion. If the coefficient of static friction is 0.380, what is the normal force acting on the box?

A.
6.08 newtons
B.
16.0 newtons
C.
16.4 newtons
D.
42.1 newtons

1 answer

To find the normal force acting on the box, we can use the relationship between the force of static friction and the normal force, which is given by the equation:

\[ F_{\text{friction}} = \mu_s \times F_{\text{normal}} \]

where \( F_{\text{friction}} \) is the force required to overcome static friction, \( \mu_s \) is the coefficient of static friction, and \( F_{\text{normal}} \) is the normal force.

Given:

  • \( F_{\text{friction}} = 16.0 \) newtons,
  • \( \mu_s = 0.380 \).

Rearranging the equation to solve for \( F_{\text{normal}} \):

\[ F_{\text{normal}} = \frac{F_{\text{friction}}}{\mu_s} \]

Now substituting in the given values:

\[ F_{\text{normal}} = \frac{16.0 , \text{N}}{0.380} \]

Calculating that gives:

\[ F_{\text{normal}} \approx \frac{16.0}{0.380} \approx 42.1 , \text{N} \]

Therefore, the correct answer is:

D. 42.1 newtons.