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In contrast, when a force exerted on the system has a component in the direction of motion, such as in [link] (d), work is done—energy is transferred to the briefcase. Finally, in [link] (e), energy is transferred from the briefcase to a generator. There are two good ways to interpret this energy transfer. One interpretation is that the briefcase’s weight does work on the generator, giving it energy. The other interpretation is that the generator does negative work on the briefcase, thus removing energy from it. The drawing shows the latter, with the force from the generator upward on the briefcase, and the displacement downward. This makes θ = 180 º size 12{θ="180"°} {} , and cos 180 º = –1 size 12{"cos 180"°= +- 1} {} ; therefore, W size 12{W} {} is negative.

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 size 12{"75" "." 0" N"} {} at an angle 35 º size 12{"35"°} {} below the horizontal and pushes the mower 25 . 0 m size 12{"25" "." 0" m"} {} on level ground? Convert the amount of work from joules to kilocalories and compare it with this person’s average daily intake of 10 , 000 kJ size 12{"10","000"" kJ"} {} (about 2400 kcal size 12{"2400"" kcal"} {} ) of food energy. One calorie (1 cal) of heat is the amount required to warm 1 g of water by 1 º C size 12{1°C} {} , and is equivalent to 4 . 184 J size 12{4 "." "184"" J"} {} , while one food calorie (1 kcal) is equivalent to 4184 J size 12{"4184"" J"} {} .

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 = Fd cos θ size 12{W= ital "Fd"" cos"θ} {} . The force, angle, and displacement are given, so that only the work W size 12{W} {} is unknown.

Solution

The equation for the work is

W = Fd cos θ . size 12{W= ital "Fd"" cos"θ} {}

Substituting the known values gives

W = 75.0 N 25.0 m cos 35.0º = 1536 J = 1.54 × 10 3 J. alignl { stack { size 12{W= left ("75" "." "0 N" right ) left ("25" "." "0 m" right )"cos " left ("35" "." 0° right )} {} #size 12{" "="1536"" J"=1 "." "54" times "10" rSup { size 8{3} } " J" "." } {} } } {}

Converting the work in joules to kilocalories yields W = ( 1536 J ) ( 1 kcal / 4184 J ) = 0 . 367 kcal size 12{W= \( "1536"`J \) \( 1`"kcal"/"4184"`J \) =0 "." "367"`"kcal"} {} . The ratio of the work done to the daily consumption is

W 2400 kcal = 1 . 53 × 10 4 . size 12{ { {W} over {"2400"`"kcal"} } =1 "." "53" times "10" rSup { size 8{ - 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.

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Section summary

  • Work is the transfer of energy by a force acting on an object as it is displaced.
  • The work W size 12{W} {} that a force F size 12{F} {} does on an object is the product of the magnitude F size 12{F} {} of the force, times the magnitude d size 12{d} {} of the displacement, times the cosine of the angle θ size 12{q} {} between them. In symbols,
    W = Fd cos θ . size 12{W= ital "Fd""cos"θ "." } {}
  • 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.

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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.

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

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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 and kilocalories.

3 . 00  J = 7 . 17 × 10 4  kcal alignl { stack { size 12{3 "." "00"" J"={}} {} #size 12{7 "." "17" times "10" rSup { size 8{ - 4} } " kcal"} {} } } {}
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A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task.

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(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.

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Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.0 gal of gasoline. Only 30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (See [link] for the energy content of gasoline.) (a) What is the magnitude of the force exerted to keep the car moving at constant speed? (b) If the required force is directly proportional to speed, how many gallons will be used to drive 108 km at a speed of 28.0 m/s?

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Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20 . 0 º size 12{"20" "." 0°} {} with the horizontal. (See [link] .) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp.

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.
3 . 14 × 10 3 J size 12{3 "." "14" times "10" rSup { size 8{3} } " J"} {}
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How much work is done by the boy pulling his sister 30.0 m in a wagon as shown in [link] ? Assume no friction acts on the wagon.

A child is sitting inside a wagon and being pulled by a boy with a force F at an angle thirty degrees upward from the horizontal. F is equal to fifty newtons, the displacement vector d is horizontal in the direction of motion. The magnitude of d is thirty meters.
The boy does work on the system of the wagon and the child when he pulls them as shown.
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A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction 25 . 0 º size 12{"25" "." 0°} {} below the horizontal. (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, using energy considerations. (e) What is the total work done on the cart?

(a) 700 J size 12{ - "700"`J} {}

(b) 0

(c) 700 J

(d) 38.6 N

(e) 0

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Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 90.0 kg, down a 60 . 0 º size 12{"60" "." 0°} {} slope at constant speed, as shown in [link] . The coefficient of friction between the sled and the snow is 0.100. (a) How much work is done by friction as the sled moves 30.0 m along the hill? (b) How much work is done by the rope on the sled in this distance? (c) What is the work done by the gravitational force on the sled? (d) What is the total work done?

A person on a rescue sled is shown being pulled up a slope. The slope makes an angle of sixty degrees from the horizontal. The weight of the person is shown by vector w acting vertically downward. The tension in the rope depicted by vector T is along the incline in the upward direction; vector f depicting frictional force is also acting in the same direction.
A rescue sled and victim are lowered down a steep slope.
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Questions & Answers

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Calculate the final velocity attained, when a ball is given a velocity of 2.5m/s, acceleration of 0.67m/s² and reaches its point in 10s. Good luck!!!
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Doc
vf=vi+at vf=2.5+ 0.67*10 vf= 2.5 + 6.7 vf = 9.2
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Two bodies P and Quarter each of mass 1000g. Moved in the same direction with speed of 10m/s and 20m/s respectively. Calculate the impulse of P and Q obeying newton's 3rd law of motion
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A disturbance that travel from one medium to another and without causing permanent change to its displacement
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In physics, a wave is a disturbance that transfers energy through matter or space, with little or no associated mass transport (Mass transfer). ... There are two main types ofwaves: mechanical and electromagnetic. Mechanicalwaves propagate through a physical matter, whose substance is being deformed
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31. Calculate the initial (from rest) acceleration of a proton in a 5.00×106 N/C electric field (such as created by a research Van de Graaff). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.
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the rubbing of one object or surface against another.
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momentum is the product of mass and it's velocity.
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Friction is a force that exist between two objects in contact. e.g. friction between road and car tires.
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Practice Key Terms 3

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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