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# Transformation of Potential and Kinetic Energy Work by rt3463df

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```									                            Work
PS-6.3 Explain work in terms of the relationship among the force
applied to an object, the displacement of the object, and the energy
transferred to the object.

PS-6.4 Use the formula W = Fd to solve problems related to work
done on an object.
Work
• The product of:
– the force applied to an object and
– the displacement the object is moved, in the
direction of the force

• Work, force, and displacement are
quantities that have magnitude and
direction
What do
you
think?
Work
• In order to do work on an object…
– A force must be applied to the object
– The object must move in the direction of the
force (displacement)
Work and Energy
• When work is done on an object, energy is
transferred to that object.
– Work is equal to the change in energy
– When a net force is applied to an object and it
moves, the work is transformed to kinetic
energy
• The kinetic energy will be greater if
– A greater force is added
– If it is applied over a greater distance
Which takes more work?
• Suppose the ramp and staircase was the
same height. Would it take more work to
lift a ball to the top or to roll it up the ramp
to the top?
Height
• If an object is lifted to some height, it gains
gravitational potential energy equal to the
work done against gravity to lift it to that
height
– The work done against gravity is the same
whether the object was lifted straight up or
rolled up a ramp
– The greater the height, the more gravitational
potential energy the object has
Work Formula
• Work = Force * displacement
• W = Fd
– Work is measured in Joules (J)
• F=W/d
– Force is measured in Newton’s (N)
• d=W/F
– Displacement is measured in meters (m)
W

F       d
Example 1
• A fork lift moves 34m carrying a 1023N
box across the warehouse floor. How
much work is done by the fork lift?

W = Fd
W = 1023 N * 34 m
W = 34,782 J
Example 2
• How much work is done by a person who
uses a force of 27.5N to move a grocery
buggy 12.3m?

W = Fd
W = 27.5 N * 12.3 m
W = 338.25 J
Example 3
• You and 3 friends apply a combined force
of 489.5N to push a piano. The amount of
work done is 1762.2J. What distance did
the piano move?

d=W/F
d = 1762.2 J / 489.5 N
d = 3.6 m
Example 4
• 55, 000J of work is done to move a rock
25m. How much force was applied?

F=W/d
F = 55,000 J / 25 m
F = 2,200 N
Practice
1. Calculate the amount of work done when moving a 567N crate a distance of 20
meters.

2. A fallen tree is lifted 2.75 meters using a force of 1565N. How much work is
done?

3. If it took a bulldozer 567.6 joules of work to push a mound of dirt 30.5 meters,
how much force did the bulldozer have to apply?

4. A frontend loader needed to apply 137 Newton's of force to lift a rock. A total of
223 joules of work was done. How far was the rock lifted?

5. A young boy applied a force of 2,550 Newton's on his St. Bernard dog who is
sitting on the boy's tennis shoes. He was unable to move the dog. How much
work did he do trying to push the dog?

6. If it takes 68 joules of work to push a desk chair across a floor 12m, what force
would be needed?

7. If a long distance runner with a weight of 596.82 newtons does 35,674.7 joules
of work during a portion of a race, what distance will she cover during that
portion?

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