15. Calculate The coefficient of kinetic friction be- tween a large box and the floor is 0.21. A person pushes
horizontally on the box with a force of 160 N for a
distance of 2.3 m. If the mass of the box is 72 kg, what is
the total work done on the box?

Wt = 26.84 [N]

Explanation:

In order to solve this problem we must use the definition of work in physics. Which tells us that this is equal to the product of force by distance.

In this case, we must sum the works of the force applied by the box and the friction force that also acts on the box.

The friction force is defined as the product of the normal force by the coefficient of friction.

f = N*μ

where:

N = normal force = m*g [N] (units of Newtons)

m = mass = 72 [kg]

g = gravity acceleration = 9.81 [m/s²]

f = friction force [N]

μ = friction coefficient = 0.21

f = 72*9.81*0.21

f = 148.32 [N]

Now the total work:

Wt = WF - Wf

where:

Wt = total work [J] (units of Joules)

WF = work by the pushing force [J]

Wf = work done by the friction force [J]

Wt = (160*2.3) - (148.32*2.3)

Wt = 26.84 [N]

Note: The friction force exerts a negative work, because this force is acting in opposite direction to the movement, therefore the negative sign.

(75°f − 32) × 5/9 = 23.889°c

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