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Answer :
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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