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Work can be positive or negative. In [link] work done lifting the mass is positive because both force and displacement are in the same direction. Likewise, when the mass is lowered the work done is negative because the force and displacement are in opposite directions. We will soon see that positive work adds energy to the system and negative work removes energy from a system.

Calculating work

Work and energy have the same units. From the definition of work, we see that those units are force times distance. Thus, in SI units, work and energy are measured in newton-meters . A newton-meter is given the special name joule    (J), and 1 J = 1 N m = 1 kg m 2 /s 2 size 12{1" J"=1" N" cdot m=1" kg" cdot m rSup { size 8{2} } "/s" rSup { size 8{2} } } {} . One joule is not a large amount of energy; it would lift a small 100-gram apple a distance of about 1 meter.

Calculating the work you do to push a lawn mower across a large lawn

How much work is done on the lawn mower by the person in [link] (a) if he exerts a constant force of 75.0 N and pushes the mower 25.0 m on level ground? Compare it with this person’s average daily intake of 10,000 kJ (about 2400 kcal) of food energy.

Strategy

We can solve this problem by substituting the given values into the definition of work done on a system, stated in the equation W = F d . The force and displacement are given, so that only the work W is unknown. Note that force and displacement are in the same direction.

Solution

The equation for the work is

W = F d .

Substituting the known values gives

W = ( 75.0 N ) ( 25.0 m ) = 1875 J = 1.88 × 10 3 J .
The ratio of the work done to the daily consumption is

W 1.00 × 10 4 k J = 1.88 × 10 3 J 1.00 × 10 7 J = 1.88 × 10 4 .

Discussion

This ratio is a tiny fraction of what the person consumes, but it is typical. Very little of the energy released in the consumption of food is used to do work. Even when we “work” all day long, less than 10% of our food energy intake is used to do work and more than 90% is converted to thermal energy or stored as chemical energy in fat.

Calculating the work done lifting a mass

The 60.0 kg mass shown in [link] is raised at a constant speed through a vertical distance of 0.800 m. What is the force that must be exerted? How much work is done lifting the mass?

Strategy

Constant speed tells us that the upward force must have the same magnitude as the downward force of gravity F = m g (to have a zero net force). Using that information, along with the given values for mass and displacement, we can determine work using W = F d .

Solution

The force is given by

F = m g = ( 60.0 k g ) ( 9.80 m / s 2 ) = 588 N .

The work done is

W = F d = ( 588 N ) ( 0.800 m ) = 470 J .

Discussion

Note that the work done lowering the mass back to its starting position would be - 407 J because force and displacement point in opposite directions. The total work done raising and lowering the mass is 407 J - 407 J = 0 J.

Section summary

  • Work is the transfer of energy by a force acting on an object as it is displaced.
  • The work W that a force F does on an object is the product of the magnitude F of the force parallel to the motion, times the magnitude d of the displacement. In symbols, W = F d .
  • The SI unit for work and energy is the joule (J), where 1 J = 1 N m = 1 kg m 2 /s 2 size 12{1" J"=1" N" cdot m="1 kg" cdot m rSup { size 8{2} } "/s" rSup { size 8{2} } } {} .
  • The work done by a force is zero if the displacement is either zero or perpendicular to the force.
  • The work done is positive if the force and displacement have the same direction, and negative if they have opposite direction.

Conceptual questions

Give an example of something we think of as work in everyday circumstances that is not work in the scientific sense. Is energy transferred or changed in form in your example? If so, explain how this is accomplished without doing work.

Give an example of a situation in which there is a force and a displacement, but the force does no work. Explain why it does no work.

Describe a situation in which a force is exerted for a long time but does no work. Explain.

Problems&Exercises

How much work does a supermarket checkout attendant do on a can of soup he pushes 0.600 m horizontally with a force of 5.00 N? Express your answer in joules.

3.00 J

A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task.

(a) Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 100 N. (b) What is the work done on the lift by the gravitational force in this process? (c) What is the total work done on the lift?

(a) 5 . 92 × 10 5 J size 12{5 "." "92" times "10" rSup { size 8{5} } " J"} {}

(b) 5 . 88 × 10 5 J size 12{ - 5 "." "88" times "10" rSup { size 8{5} } " J"} {}

(c) The net force is zero.

Calculate the work done on a crate pushed 4.00 m up along a ramp (see [link] ). The man exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed.

A person is pushing a heavy crate up a ramp. The force vector F applied by the person is acting parallel to the ramp.
A man pushes a crate up a ramp.
2.00 × 10 3 J

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. (a) What is the work done on the cart by friction? (b) What is the work done on the cart by the gravitational force? (c) What is the work done on the cart by the shopper? (d) Find the force the shopper exerts parallel to the direction of motion. (e) What is the total work done on the cart?

(a) -700 J

(b) 0

(c) 700 J

(d) 35.0 N

(e) 0

Practice Key Terms 3

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Source:  OpenStax, Concepts of physics. OpenStax CNX. Aug 25, 2015 Download for free at https://legacy.cnx.org/content/col11738/1.5
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