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7-2 The work done by a constant force exerted on a moving object 295

EXAMPLE 7-2 Up the Hill
You need to push a box of supplies (weight 225 N) from your car to your campsite, a distance of 6.00 m up a 5.00 ° incline.
You exert a force of 85.0 N parallel to the incline, and a 56.0-N kinetic friction force is exerted on the box by the ground.
Calculate (a) how much work you do on the box, (b) how much work the force of gravity does on the box, (c) how much
work the friction force does on the box, and (d) the net work done on the box by all forces exerted on it as it moves up
the incline.

Set Up y
Because this problem involves forces, we draw a free-body Work done by a constant force,
diagram for the box. The forces do not all point along the displacement: F

direction of the box’s displacement ,d so we’ll have to use W = Fd cos θ (7-2) n
Equation 7-2 to calculate the work done by each force. F you x
F k d
5.00°
F g

Solve

(a) The force you exert on the box is in the same W you = F you d cos0 °
direction as the box’s displacement, so θ =°0 in = (85.0 N)(6.00m)(1)
Equation 7-2. = 5.10 10 Nm
2
×
2
×
= 5.10 10 J
(b) The angle between the gravitational force and the W grav = g F d cos 95.0°
displacement is θ = 90.0 ° + 5.00 ° = 95.0 °. Because this = (225 N)(6.00m)( 0.0872)
−
is more than 90 ,° cos θ is negative and the gravitational =− 1.18 10 J
2
×
force does negative work on the box.
(c) The friction force points opposite to the displacement, W friction = F d cos 180 °
k
so for this force θ = 180° and cos θ =− 1. = (56.0 N)(6.00m)( −1)
2
=−3.36 10 J
×
(d) The net work done on the box is the sum of the work W total = W you + W grav + W friction + W n
done by all four forces that are exerted on the box. The = (5.1010J)( × 2
2
×
+−1.18 10 J)
normal force points perpendicular to the displacement, so +− (3.3610J)0
2
+
×
0
it does zero work (for this force, θ = 90° and cos θ =°).
2
×
= 0.56 10 J = 56J
Reflect
To check our result, let’s calculate how much work is done Net force up the incline:
by the net force that is exerted on the box. Our discussion ∑ F x = F you − F − g F sin5.00°
of Equation 7-2 tells us that we need only the component = 85.0N − k 56.0N − (225 N)(0.0872)
of the net force in the direction of the displacement, which
in our figure is the x component. So the work done by the = 9.4N
∑
net force is just F x multiplied by d. (positive, so the net force is uphill)
There are two observations we can use to ensure this Work done by net force:
makes sense. First, the work done by the net force has the = ∑ (
same value as the sum of the work done by the individual W net F x ) d
forces. It has to, because this is just another way to describe = (9.4 N)(6.00m)
the exact same situation! Second, the net force is in the = 56 J
direction of the displacement, so the box picks up speed as This equals the sum of the work done by the four forces
it moves and the net work done on the box is positive. individually.
Extend: We’ll use both of these observations in the next section to relate the net work done on an object to the change in
its speed.
NOW WORK Problem 3 from The Takeaway 7-2.

Uncorrected proofs have been used in this sample. Copyright © Bedford, Freeman & Worth Publishers.
Distributed by Bedford, Freeman & Worth Publishers. For review purposes only. Not for redistribution.

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