As shown in the figure below, a box of mass m = 65.0 kg initially at rest s push
ID: 1584520 • Letter: A
Question
As shown in the figure below, a box of mass m = 65.0 kg initially at rest s pushed a distance d 830 m across a frictionless warehouse floor under,the influence of two forces. A force of F1 215 N directed at an angle of 30.0° below the horizontal and a force F2 110 N acting in a direction that opposes the motion. Determine the following 30° (a) work done on the box by the force F (b) work done on the box by the force F2 (e) work done on the box by the force of gravity F (d) work done on the box by the normal force N (e) net work done on the box by finding the sum of all the work done on the box by the individual forces (f) net work done on the box by finding the net force acting on the box and then finding the work done by this net forceExplanation / Answer
work done = F*d*costheta
theta = angle between force and displacement
(a)
work done on the box by the force F1 = F1*d*cos30 = 215*83*cos30 = 15454.22 J
(b)
theta = 180
work done on the box by the force F2 = F1*d*cos180 = 110*83*cos180 = -9130 J
(c)
Fg is verticallay downwards
work done on the box by the force of gravity Fg = Fg*d*cos90 = 0
(d)
N is verticallay upwards
work done on the box by the normal force N = N*d*cos90 = 0
(e)
net work done = 15454.22 - 9130 = 6324.22 J
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