91Ó°ÊÓ

The given data can be listed below as:

• The acceleration of the elevator is$a=1.90m/s^2$ .
• The total mass of the contents and the box is$M=36.0kg$ .
• The coefficient of the kinetic friction between the elevator floor and the box is${\mathrm{μ}}_k=0.32$ .

The friction is described as the force that mainly resists the motion of an object. The frictional force is equal to the product of the coefficient of friction and the normal force.

The equation of the frictional force is expressed as:

$f_k={\mathrm{μ}}_{k}N$ …(¾±)

Here,$f_k$ is the frictional force,${\mathrm{μ}}_k$ is the coefficient of kinetic friction and N is the normal force.

As the elevator is moving in the upward direction, the equation of the total mass of the box and its contents is expressed as:

$$\begin{gathered} N-mg=ma \\ N=m\left(g+a\right) \end{gathered}$$

Here, m is the total mass of the contents and g the box and is the acceleration due to gravity.

Substitute$m\left(g+a\right)$ for N in the equation (i).

$f_k={\mathrm{μ}}_{k}m\left(g+a\right)$

As the computer is needed to move with constant speed, then the acceleration of the computer will be zero. Hence, the force exerted and the frictional force should be the same.

The equation of the force exerted is expressed as:

$F={\mathrm{μ}}_{k}m\left(g+a\right)$

Here, F is the force exerted.

Substitute the values in the above equation.

$$\begin{gathered} F=\left(0.32\right)\left(36kg\right)\left(9.8m/s^2+1.9m/s^2\right) \\ =\left(11.52kg\right)\left(11.7m/s^2\right) \\ =134.9kg×m/s^2×\frac{1N}{1kg×m/s^2} \\ =134.9N \end{gathered}$$

Thus, the magnitude of the force that must be applied is 134.9 N .