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A pickup truck is traveling with constant velocity up an interstate entrance ramp with a 5.50° slope. A 50.0-kg box sits on the bed of the truck and is also resting against the tail-gate of the truck. The box does not move relative to the bed of the truck. Assume that there is negligible friction between the bed of the truck and the box. During an interval in which the truck travels 395 m up the ramp entrance, what is the net work done on the box?

Answer :

fabianb4235

Answer:

W=18569.87Joules

Explanation:

Hello!

To solve this problem we must remember what work is and how it is calculated.

Remember that Work is the energy needed to move a body

The work of a body is equal to the product of the force in a direction parallel to the displacement by the distance traveled.

hen to solve the problem we draw a free-body diagram (see attached image) of the box and find the force due to the weight of the box that is parallel to the direction of the ramp, finally, we multiply this force by the distance (395m )

w=WORK

f=force

d=distance=395m

m=mass(50kg)

g=gravity

[tex]W=Fd\\W=mgsen5.5d\\W=(50kg)(9.81m/s^2)(sen5.5)(395m)=18569.87Joules[/tex]

W=18569.87Joules

${teks-lihat-gambar} fabianb4235

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