As shown in the figure below, a box of mass m = 58.0 kg (initially at rest) is p
ID: 1336122 • Letter: A
Question
As shown in the figure below, a box of mass m = 58.0 kg (initially at rest) is pushed a distance d = 73.0 m across a rough warehouse floor by an applied force of Fa = 236 N directed at an angle of 30.0 degree 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 WA = 14919 J work done by the force of gravity Vig = work done by the normal force WN =0 work done by the force of friction Wf = Pay careful attention to the angle between the direction of the force of friction and the displacement vector. Do you recall how to determine the force of friction in terms of the normal force and the coefficient of friction? A good diagram showing all the forces and their components will help you determine a correct expression for the normal force needed in order to determine the force of friction. J Calculate the net work on the box by finding the sum of all the works done by each individual force. WNet = Now find the net work by first finding the net force on the box, then finding the work done by this net force.Net = JExplanation / Answer
Here ,
Normal force , N = mg + FA * cos(theta)
N = 58 * 9.8 + 236 * cos(36)
N = 772.8 N
d)
work done by friction = - frictional force * d
work done by friction = -u * N * d
work done by friction = - 0.1 * 778.3 * 73
work done by friction = -5641.3 J
the work done by friction is -5641.3 J
e)
Net work done = 14919 -5641.3 J
Net work done = 9277 J
the Net work done on the box is 9277 J
f)
net force on the block ,
Fnet = 236 * cos(30) - 0.1 * 772.8
Fnet = 127N
net work done = Fnet*distance
net work done = 127* 73
net work done = 9277 J
the net work done is9277 J
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