As shown in the figure below, a box of mass m = 59.0 kg (initially at rest) is pushed a distance d = 80.0 m across a rough warehouse floor by an applied force of FA = 220 N directed at an angle of 30.0° beld 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.) 30° (a) work done by the applied force WA = J (b) work done by the force of gravity W₂ = (c) work done by the normal force WN - rough surface wwwwwwwwwwwww •dy

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As shown in the figure below, a box of mass m = 59.0 kg (initially at rest) is pushed a distance d = 80.0 m across a rough warehouse floor by an applied force of FA = 220 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 =
(b) work done by the force of gravity
Wg
=
(c) work done by the normal force
WN =
(d) work done by the force of friction
Wf =
rough surface
www
(e) Calculate the net work on the box by finding the sum of all the works done by each individual force.
W Net =
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.
W Net
=
Transcribed Image Text:As shown in the figure below, a box of mass m = 59.0 kg (initially at rest) is pushed a distance d = 80.0 m across a rough warehouse floor by an applied force of FA = 220 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 = (b) work done by the force of gravity Wg = (c) work done by the normal force WN = (d) work done by the force of friction Wf = rough surface www (e) Calculate the net work on the box by finding the sum of all the works done by each individual force. W Net = 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. W Net =
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