A truck travels uphill with constant velocity on a highway with a 7.50° slope. A
ID: 1779181 • Letter: A
Question
A truck travels uphill with constant velocity on a highway with a 7.50° slope. A 40.0-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 356 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
work done is the scalar product of force and displacement, or the displacement in the direction of force times force of you consider force component along the displacement times displacement
normal force is perpendicular to displacement hence zero work dont by normal force
weight component acting on the 50 kg package is 50 g sin 6.5
frictional force will be equal and opposite to this because the truck moves with zero acceleration
so net force has to be zero since acceleration is zero net force = ma
net work done is hence zero
work done by gravity is= 40 g sin 7.5 * 356=18215 joule
work done by friction is= - 40 g sin 7.5 * 356 = -18215 joule
workdone by normal force = 40gcos7.5 = 388.64 joule.
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