Work
The Flower Carrier, Diego Rivera, 1935
What is work, in the science-sense?
Pull a book off the shelf.
Pull yourself up exercising.
Push a box up a ramp.
What do all these have in common?
1: A force is applied...
2: ...over a distance.
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Work is done when a constant force is applied in the direction of displacement.
W = Fǁd
Example
How much work is done if you exert 6.67 N of force to lift a book through a vertical distance of 0.750 m?
W = Fǁd
= 6.67N x 0.750m
= 5.00 J
Example
Forces applied at an angle
Only the force applied in the direction of the displacement does work.
W = Fǁd
If the force is applied at an angle to the moving object only the component in the direction of the motion is doing work.
W = Fdcosθ
θ is the angle of the force from the direction of motion
(math note: this is the dot product of F and d which is why work is scalar)
θ
Example
A small plane tows a glider at constant speed and altitude. If the plane does 2.00x105 J of work to tow the glider 145 m and the tension in the tow rope is 2560 N, what is the angle between the two rope and the horizontal?
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Example
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You want to load a 35 kg box into the back of a van. One way is to lift the box straight up 1.00 m to the back of the van. Alternatively, you can slide the box up a 19.47o (from ground) ramp with wheels (negligible friction) a distance of 3.00 m. Which method would take more work at constant speed? (work out on board)
They would require the same work! (342 J)
One requires more force and one requires more distance.
Some important notes about work
1) Work has a sign that corresponds to the direction of the force relative to the motion of the object.
2) The total work done on an object is the sum of the work done by all forces acting on the object
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Example
Wanda Wren pulls her suitcase at constant velocity through the airport on level ground with a force of 35.0 N at 55.0o above the horizontal for a distance of 30.0 m.
Example
Wanda Wren pulls her suitcase at constant velocity through the airport on level ground with a force of 35.0 N at 55.0o above the horizontal for a distance of 30.0 m.
a)
b) i) WN = Wg = 0 J because they are perpendicular to the motion of the suitcase
ii) Wpull = (35N)(30m)(cos 55o) = 602 J
(+ because same direction as motion)
iii) Wf = (35N)(30m)(cos180o) = -602 J
(- because opposite direction, 180o, from motion)
c) Wtot = 0J + 0J + 602 J + -602 J = 0 J
Fpull
Fg
FN
Ff