Answer to Question #150227 in Mechanics | Relativity for Justin Vidal

Question

A shopper pushes her 25 kg grocery cart by a force of 225 N inclined at an
angle of 60° with the horizontal through a distance of 7.5 m as shown in the
free body diagram. Find the work done by (a) the 225 N force and (b) friction.
Assume that the coefficient of kinetic friction is 0.32.

Answer 1

Solution:

a)

Work = the Force in line with the motion times the distance.

$Fcos60*7.5= \dfrac{15F}{4} =843.75J$

b)

Force of friction:

$F_f=(225*sin(60)+25*9.8)*0.32=140.8N$

F_f is in the opposite direction of motion so that:

Work of friction =$-140.8*7.5=-1056J$