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  • Explain how an object must be displaced for a force on it to do work.
  • Explain how relative directions of force and displacement determine whether the work done is positive, negative, or zero.

What it means to do work

The scientific definition of work differs in some ways from its everyday meaning. Certain things we think of as hard work, such as writing an exam or carrying a heavy load on level ground, are not work as defined by a scientist. The scientific definition of work reveals its relationship to energy—whenever work is done, energy is transferred.

For work, in the scientific sense, to be done, a force must be exerted and there must be displacement in the direction of the force.

Formally, the work    done on a system by a constant force is defined to be the product of the component of the force in the direction of motion times the distance through which the force acts . For one-way motion in one dimension, this is expressed in equation form as

W = F cos θ d , size 12{W= lline F rline left ("cos"θ right ) lline d rline } {}

where W size 12{W} {} is work, d size 12{d} {} is the displacement of the system, and θ size 12{θ} {} is the angle between the force vector F size 12{F} {} and the displacement vector d size 12{d} {} , as in [link] . We can also write this as

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

To find the work done on a system that undergoes motion that is not one-way or that is in two or three dimensions, we divide the motion into one-way one-dimensional segments and add up the work done over each segment.

What is work?

The work done on a system by a constant force is the product of the component of the force in the direction of motion times the distance through which the force acts . For one-way motion in one dimension, this is expressed in equation form as

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

where W size 12{W} {} is work, F size 12{F} {} is the magnitude of the force on the system, d size 12{d} {} is the magnitude of the displacement of the system, and θ size 12{q} {} is the angle between the force vector F size 12{F} {} and the displacement vector d size 12{d} {} .

Five drawings labeled a through e. In (a), person pushing a lawn mower with a force F. Force is represented by a vector making an angle theta with the horizontal and displacement of the mower is represented by vector d. The component of vector F along vector d is F cosine theta. Work done by the person W is equal to F d cosine theta. (b) A person is standing with a briefcase in his hand. The force F shown by a vector arrow pointing upwards starting from the handle of briefcase and the displacement d is equal to zero. (c) A person is walking holding the briefcase in his hand. Force vector F is in the vertical direction starting from the handle of briefcase and displacement vector d is in horizontal direction starting from the same point as vector F. The angle between F and d theta is equal to 90 degrees. Cosine theta is equal to zero. (d) A briefcase is shown in front of a set of stairs. A vector d starting from the first stair points along the incline of the stair and a force vector F is in vertical direction starting from the same point as vector d. The angle between them is theta. A component of vector F along vector d is F d cosine theta. (e) A briefcase is shown lowered vertically down from an electric generator. The displacement vector d points downwards and force vector F points upwards acting on the briefcase.
Examples of work. (a) The work done by the force F size 12{F} {} on this lawn mower is Fd cos θ size 12{ ital "Fd""cos"θ} {} . Note that F cos θ size 12{F"cos"θ} {} is the component of the force in the direction of motion. (b) A person holding a briefcase does no work on it, because there is no displacement. No energy is transferred to or from the briefcase. (c) The person moving the briefcase horizontally at a constant speed does no work on it, and transfers no energy to it. (d) Work is done on the briefcase by carrying it up stairs at constant speed, because there is necessarily a component of force F size 12{F} {} in the direction of the motion. Energy is transferred to the briefcase and could in turn be used to do work. (e) When the briefcase is lowered, energy is transferred out of the briefcase and into an electric generator. Here the work done on the briefcase by the generator is negative, removing energy from the briefcase, because F size 12{F} {} and d size 12{d} {} are in opposite directions.

To examine what the definition of work means, let us consider the other situations shown in [link] . The person holding the briefcase in [link] (b) does no work, for example. Here d = 0 size 12{d=0} {} , so W = 0 size 12{W=0} {} . Why is it you get tired just holding a load? The answer is that your muscles are doing work against one another, but they are doing no work on the system of interest (the “briefcase-Earth system”—see Gravitational Potential Energy for more details). There must be displacement for work to be done, and there must be a component of the force in the direction of the motion. For example, the person carrying the briefcase on level ground in [link] (c) does no work on it, because the force is perpendicular to the motion. That is, cos 90 º = 0 size 12{"cos""90""°="0} {} , and so W = 0 size 12{W=0} {} .

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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Can you compute that for me. Ty
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what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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Adjanou
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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answer
Magreth
progressive wave
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Physics 110 at une. OpenStax CNX. Aug 29, 2013 Download for free at http://legacy.cnx.org/content/col11566/1.1
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