Integrating Force to find Work
Work, measured in either joules(Newton-meters) or footpounds, is the product of the force necessary to move and
object and the displacement or distance of the object caused
by the force. It is given by the equation W = Fd
If a 10 lb block is pushed 4 feet across a floor, the work done
in pushing the block would be 40 foot-lbs.
If a 20 kg block is pushed 3 meters across a floor, the work
done in pushing the block would be
W = Fd = (20 kg 9.8 m/s2)(3 meters) = 588
Newton-meters.
The difference here is that with
metric weight measurements, we
have to multiply by 9.8 m/s2
Suppose that the force required to move an object x meters across
the floor decreases as the object gains speed so that the force can
be given by the function F(x) =Ce0.43x.
If the force required to budge the object from rest is 49 N, find C.
At any point x, the force required is 49e0.43x and the over a
small distance x the work done would be W = 49e0.43x  x.
Over a distance from 0 to 6 meters we can approximate it by
adding all the small x partitions so that we get
k

49 e
 0.43 x
 x
n1
As k goes to infinity, this summation becomes
6
 4 9 e
0
 0 .4 3 x
dx
 105.32 joules (or Newton-meters)
Work 
b
 a F ( x ) dx
…in which F(x) = a force function
…but there are cases in which dx
can be part of the force or weight
of an object being moved…
Promise
Tell
Youusknow
what
or we’ll
what
we want
leave
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Muhahahahahahaaaa
you down
towant…
hear…
there…
Tina and Danielle Yee have Mr. Murphy (weighing 82 kg) dangling from a rope, 25 m
long and weighing 5 kg/m , off the edge of a building. When Mr. Murphy finally
promises a lifetime supply of cookies on demand, they hoist him back up to the top
of the building. How much total work is done by Tina and Danielle raising both the
rope and Mr. Murphy to the top of the building?
The Mr. Murphy part is easy…
W = F  d = weight  distance
W = F  d = 82 kg (9.8 m/s2) 25 m
W = 20,090 j
Ellie and Lisa have Mr. Murphy (weighing 82 kg) dangling from a rope, 25 m long
and weighing 5 kg/m, off the edge of a building. When Mr. Murphy finally
promises to bake them a lifetime supply of cookies, they hoist him back up to the
top of the building. How much total work is done by Ellie and Lisa raising both the
rope and Mr. Murphy to the top of the building?
The rope part is a little different…
x
dx
A small piece of the rope with length dx cm
…and weighing 5 kg/m  dx m
…is pulled up by a distance of x meters
Tina and Danielle (Yee, of course) have Mr. Murphy (weighing 82 kg) dangling from
a rope, 25 m long and weighing 5 kg/m, off the edge of a building. When Mr.
Murphy finally promises to bake them a lifetime supply of cookies, they hoist him
back up to the top of the building. How much total work is done by Tina and
Danielle raising both the rope and Mr. Murphy to the top of the building?
The rope part is a little different…
Work (for one small piece of rope) = Weight  Distance
Don’t forget
about gravity
5 kg/m  dx m
Work  9 . 8 
25
x meters
5 x dx =15,312.5 j
0
The sum of all
Distance
the little
Force (or in this case the weight)
pieces of
of one little piece of the rope
rope