Question

a boy pulls 23 kg box with a 100 N force at 39 above a horizontal surface if the coefficient of kinetic friciton between the box and the horizontal surface is 0.16 and the box is pulled a distnace of 34.0 what is the net work done on the box?

Answer 1

Answer:

Total work done is 2606.08 J.

Explanation:

Given :

Mass of box , m = 23 kg .

Force applied , F = 100 N .

Angle from horizon , $\theta=39^o$.

Coefficient of kinetic friction , $\mu=0.16$.

Distance travelled by box , d = 34 m .

Now ,

Total work done = work done by boy + work done by friction.

$W=Fdcos\theta+f_sdcos\theta=Fd-\mu(mg) \\\\W=100\times cos39^o\times 34-0.16\times 23 \times 9.8 \\\\W=77.71\times 34-36.06=2606.08\ J.$

Hence , this is the required solution.