A truck travels uphill with constant velocity on a highway with a 6.0° slope. A

Question

A truck travels uphill with constant velocity on a highway with a 6.0° slope. A 55-kg package sits on the floor of the back of the truck and does not slide, due to a static frictional force. During an interval in which the truck travels 315 m, find the following.

(a) What is the net work done on the package?
____J

(b) What is the work done on the package by the force of gravity?
____J

(c) What is the work done on the package by the normal force?
____J

(d) What is the work done on the package by the friction force?
____J

Explanation / Answer

truck moving with constant velocity => a = 0

now package does not slide

=> mgsin(theta) = Us*mgcos(theta)

a) net workdone on package = 0 J

b) workdone by force of gravity = mgsin(theta)*(-315) = 55*9.8*sin(6 degrees)*(-315) = - 17.74 kJ

c) workdone by normal force = 0 because no displacement in normal direction.

d)  workdone by force of friction = Us*mgcos(theta)*(315) = mgsin( theta)*(315) = 17.74 kJ

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