Jack pushes an 8.0kg box is through a distance of 2.2m up a 20o frictionless inc
ID: 2177672 • Letter: J
Question
Jack pushes an 8.0kg box is through a distance of 2.2m 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)Normal force and distance moved are perpendicular to each other. So work done = 0 b)Work Done by gravity = 8*9.81*SIN(RADIANS(20))*2.2 = 59.052 J c)Work done by Jack = 8*9.81*SIN(RADIANS(20))*2.2 = 59.052 j
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