12.1 Please help, last 3 chegg experts have got (e) and (f) answers wrong. As sh

Question

12.1

Please help, last 3 chegg experts have got (e) and (f) answers wrong.

As shown in the figure below, a box of mass m = 66.0 kg (initially at rest) is pushed a distance d = 99.0 m across a rough warehouse floor by an applied force of F_a = 224 N directed at an angle of 30.0degree below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.100. Determine the following. (For parts (a) through (d), give your answer to the nearest multiple of 10.) work done by the applied force work done by the force of gravity work done by the normal force work done by the force of friction Calculate the net work on the box by finding the sum of all the works done by each individual force. Knowing the work done on the box by each force acting on the box, we can determine the net work done on the box. J Now find the net work by first finding the net force on the box, then finding the work done by this net force. A good diagram showing all the forces and their components will help you determine the correct expression for the net force. Knowing the net force acting on the box and the distance this force acts, how can we determine the net work done on the box? Pay careful attention to the angle between the direction of the net force acting on the box and the displacement vector. J

Explanation / Answer

(e) Net work done by finding sum of each work
= Workdone by applied force + work done by friction
= 19205 - 7518.654 = 11686.346 J
(f) Now we have to calculate the net force
Applied force = 224 Cos30 = 193.9896 N
Friction force = uN where u is coefficient of friction and N is the normal reaction
Friction force = 0.1*(224Sin30 + (66*9.81)) = 75.946 N
Net force = 193.9896 - 75.946 = 118.0436 N
Work done = net force*distance = 11686.3164 J

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