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Roy pushes a box of mass 17.5 kilograms across a floor by exerting a horizontal force of 97.9 newtons. The coefficient of kinetic friction between the box and the floor is 0.369, and Roy pushes the box 15.3 meters. What work does friction do? Include units in your answer. Answer must be in 3 significant digits.

Respuesta :

Consider that the work done by the friction force is given by:

[tex]W_r=F_r\cdot d=\mu_kN\cdot d=\mu_k\cdot m\cdot g\cdot d[/tex]

where,

μk: coefficient of kinetic friction = 0.369

Fr: friction force

N: normal force = m*g

m: mass of the box = 17.5 kg

g: gravitational acceleration constant = 9.8m/s^2

d: distance = 15.3 m

Replace the previous values of the parameters into the formula for Wr:

[tex]\begin{gathered} W_r=(0.369)(17.5kg)(9.8\frac{m}{s^2})(15.3m) \\ W_r=968.23755\approx968J \end{gathered}[/tex]

Hence, the work done by the friction force is approsimately 968J

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