Question

Roy pushes a box of mass 16.6 kilograms across a floor by exerting a horizontal force of 99.1 newtons. The coefficient of kinetic friction between the box and the floor is 0.324, and Roy pushes the box 11.4 meters. What is the net work done on the box? Include units in your answer. Answer must be in 3 significant digits.

Answer 1

529 Joules

Step-by-step explanation:

Let us remind ourselves of the definition of work. if a force F is acting on a body and it produces a displacemnt of Δx, then the work done is

`W=F\Delta x`

Now, in our case, what are the forces acting on the box?

The force is the sum of the frictional and the force exerted by the boy. These forces are shown in the sketch below.

Therefore, the net force acting on the box is

`F=F_{\text{boy}}-F_{\text{friction}}`

where F_boy = 99.1 N.

Now, the force of friction is

`F_{\text{friction}}=\mu mg`

where μ = coeffient of friction = 0.324, m = mass = 16.6 kg, and g = acceleration due to gravity = 9.8m/s^2.

Putting these values into the above equation gives

`F_{\text{friction}}=(0.324)(16.6)(9.8)=52.71N`

Therefore,

`F=F_{\text{boy}}-F_{\text{friction}}=99.1-52.71=46.39N`

Therefore, the work done is

`W=F\Delta x=46.39\cdot11.4=528.865J`

rounding to the three significant digits gives

`\boxed{W=520J\text{.}}`

which is our answer!

Hence the work done on the box is 529 Joules.