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As shown in the figure below, a box of mass m = 50.0 kg (initially at rest) is p

ID: 1283442 • Letter: A

Question

As shown in the figure below, a box of mass m = 50.0 kg (initially at rest) is pushed a distance d = 71.0 m across a rough warehouse floor by an applied force of FA = 222 N directed at an angle of 30.0 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.)

(a) work done by the applied force

WA = J



(b) work done by the force of gravity

Wg = J



(c) work done by the normal force

WN = J



(d) work done by the force of friction

Wf = J



(e) Calculate the net work on the box by finding the sum of all the works done by each individual force.

WNet = J



(f) Now find the net work by first finding the net force on the box, then finding the work done by this net force.

WNet = J

As shown in the figure below, a box of mass m = 50.0 kg (initially at rest) is pushed a distance d = 71.0 m across a rough warehouse floor by an applied force of FA = 222 N directed at an angle of 30.0½ 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.) (a) work done by the applied force WA = J (b) work done by the force of gravity Wg = J (c) work done by the normal force WN = J (d) work done by the force of friction Wf = J (e) Calculate the net work on the box by finding the sum of all the works done by each individual force. WNet = J (f) Now find the net work by first finding the net force on the box, then finding the work done by this net force. WNet = J

Explanation / Answer

a ) Fcos(30) * 71 meter = 13650.2924 J

b) Work done by gravity = 0 ; since there is no displacement in vertical direction

c) Work done by Normal force = 0 ; since there is no displacement in vertical direction

d) friction force = (mu) * Normal force

normal force = Mg + Fsin(30) = 601 Newton

frictional force = 601N * 0.1 = 60.1 N

Work by friction = - 60.1N * 71meters = - 4267.1 J

e) Wnet = Work by force + work by friction + work by gravity + work by Normal

= 13650.2924 - 4267.1 +0 +0 = 9383.1924 J

f) Net vertical force = 0 ; Net horizontal force = Fcos(30) - (mu) *N = 192.25764 - 60.1 = 132.15764

Worknet = 132.15764 * 71 =9383.1924 J

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