As shown in the figure below, a box of mass m = 68.0 kg (initially at rest) is p
ID: 1422507 • Letter: A
Question
As shown in the figure below, a box of mass m = 68.0 kg (initially at rest) is pushed a distance d = 92.0 m across a rough warehouse floor by an applied force of FA = 214 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 = 68.0 kg (initially at rest) is pushed a distance d = 92.0 m across a rough warehouse floor by an applied force of FA = 214 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 = 68.0 kg (initially at rest) is pushed a distance d = 92.0 m across a rough warehouse floor by an applied force of FA = 214 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
Explanation / Answer
As given in the question,
m = 68 kg, d = 92 m, FA = 214 N, = 30° and = 0.1
(a) Work done by the applied force,
WA = FA cos * d = 214 cos30° * 92 = 17050.31 J
(b) Work done by the force of gravity,
Since there is no placement in the direction of gravity force => Wg = 0
(c) Work done by the normal force,
Since there is no placement in the direction of normal force => WN = 0
(d) For the work done by the force of friction (Wf),
The normal force: N = mg + Fsin
Friction force: Ff = *N = *(mg + Fsin)
= 0.1*{ (68*9.8) + 214 sin30°} = 77.34 N
The direction of Friction force and "d" are opposite so it's work will be negative,
So, Wf = Ff*(- d) = 77.34*(- 92) = - 7115.28 J
(e) The net work on the box by finding the sum of all the works done by each individual force,
WNet = WA + Wg + WN +Wf = 17050.31 + 0 + 0 + (- 7115.28) = 9935.03 J = 9.935*10^3 kJ
(f) This will be the same exact exercise except that the distance will not be multiplied by the force until after a
net force is calculated. Of course the net force will be the Applied force horizontal component minus the friction
force.
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