Jack pushes an 8.0kg box is through a distance of 4m up a 20 o frictionless incl
ID: 1283443 • Letter: J
Question
Jack pushes an 8.0kg box is through a distance of 4m up a 20o frictionless incline. The applied force is parallel to the incline and the speed of the box is constant.
(a) What is the work done by the normal force?
J
(b) What is the work done by gravity?
J
(c) What is the work done by Jack?
J
(d) What is the magnitude of the force that Jack applied to the box? Use the concept of work to solve this one.
N
(e) Suppose that the box is moving up the hill at a constant velocity of 2.0 m/s. If the applied force you calculated in part d is doubled, calculate the speed of the box after it has traveled 5.0-m (with the newly applied force). Remember, work done by the ALL the forces acting on the box equals the change in kinetic energy.
m/s
Explanation / Answer
a) W_n = 0 J...
B) W-g = 0 J....
C) W_jack = mg*sin 20*S = 8*9.8*sin 20 *4= 107.28 J...
D) W-jack = F*S = `107.28...
F = 107.28/S = 107.28/4 = 26.82 N....
e) net force is 2*26.82 - 26.82 = 26.82.....
W = (1/2)*m*(v^2-u^2) = (1/2)*8*(v^2-2^2) ....
26.82*4 = 4*(v^2-4)....
v^2-4 = 26.82...
v^2 = 26.82+4 ....
v = 5.56 m/sec
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