8. The coefficient of static friction between a 150-kg box of medical equipment and the floor is 0.35. Would 450-N force applied horizontally be enough to cause the box to move from rest?

is the net force positive?

Net force=450-mu*150*9.8

To determine if the 450-N force applied horizontally would be enough to cause the box to move from rest, we need to compare the force with the maximum static friction force that can be generated between the box and the floor.

The formula for static friction force is:

\(f_{\text{friction}} = \mu_s \times \text{Normal force}\)

Where:
\(f_{\text{friction}}\) = static friction force
\(\mu_s\) = coefficient of static friction
\(\text{Normal force}\) = force perpendicular to the contact surface (in this case, the weight of the box)

First, let's find the normal force. The normal force is equal to the weight of the box, which can be calculated using the formula:

\( \text{Normal force} = \text{mass} \times \text{acceleration due to gravity} \)

Given that the mass of the box is 150 kg and the acceleration due to gravity is approximately 9.8 m/s², we can calculate the normal force:

\( \text{Normal force} = 150 \, \text{kg} \times 9.8 \, \text{m/s²} \)

\( \text{Normal force} = 1470 \, \text{N} \)

Now, we can calculate the maximum static friction force using the coefficient of static friction:

\( f_{\text{friction}} = 0.35 \times 1470 \, \text{N} \)

\( f_{\text{friction}} = 514.5 \, \text{N} \)

The maximum static friction force between the box and the floor is 514.5 N.

Since the applied force of 450 N is less than the maximum static friction force of 514.5 N, the box will not move from rest.