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Class 11 Physics (India)
Introduction to work review
Review the key concepts and skills for energy and work. Understand how work is the area under a force vs. displacement graph and how force and displacement produce work.
Key terms
Term (symbol) | Meaning |
---|---|
Energy (E) | Measurement of the ability to do work. SI unit is joule (start text, J, end text). |
Work (W) | Change in energy by the transfer of energy from one system to another. Scalar quantity with units of joules (start text, J, end text). |
Joules (start text, J, end text) | SI unit for energy. Applying a net force of 1, start text, N, end text to an object over a displacement of 1, start text, m, end text requires 1, start text, J, end text of energy. 1, start text, J, end text, equals, 1, start text, N, end text, dot, 1, start text, m, end text, equals, 1, start text, k, g, end text, dot, start fraction, start text, m, end text, squared, divided by, start text, s, end text, squared, end fraction |
Equations
Equation | Symbol breakdown | Meaning in words |
---|---|---|
W, equals, F, d, cosine, theta | W is work done on an object, F is the magnitude of force on the object, d is the magnitude of the object’s displacement, and theta is the angle between the vectors of force F and displacement d. | Work is the product of an object’s displacement and the component of force exerted parallel to the direction of the object’s motion. |
W, equals, delta, E | W is work and delta, E is the change of energy. | Work is the change of energy for a system. |
How to find work from a Force vs. displacement graph
The force applied to an object can be graphed as a function of the position of the object. Work is the area under the curve of the force vs. position graph. Areas above the position axis are positive work and areas below the axis are negative work. If the force is not constant, we can divide the graph into sections with simpler shapes and add up the work in each section. To find the total work done on the object in the force vs. position graph in Figure 1 over displacement d, start subscript, 1, end subscript, plus, d, start subscript, 2, end subscript, the areas of A, start subscript, 1, end subscript and A, start subscript, 2, end subscript can be added together.
A, start subscript, 1, end subscript is a rectangle of height F, start subscript, o, end subscript and width d, start subscript, 1, end subscript. A, start subscript, 2, end subscript is a triangle of height F, start subscript, o, end subscript and base d, start subscript, 2, end subscript. The total work done on the object over d, start subscript, 1, end subscript, plus, d, start subscript, 2, end subscript is
For worked examples of finding work from a force vs. displacement graph, watch our video about calculating work from force vs. displacement graphs.
Common mistakes and misconceptions
1) People forget that forces acting perpendicular to displacement do zero work. Since cosine, 90, degrees, equals, 0, no work occurs when force F and displacement d are perpendicular even though a perpendicular force can change the object’s direction of motion. For example, if a box is pulled across the level floor, the floor does zero work on the box even though the box is moving. This is because the normal force is perpendicular to the horizontal displacement of the box.
2) People forget what the sign of work means. Positive work on a system means it receives energy from its surroundings. Negative work on a system means it gave energy to its surroundings. Negative work occurs when the force has a component in the direction opposite the displacement.
Learn more
For deeper explanations of work, see our video with example problems about work.
To check your understanding and work toward mastering these concepts, check out the exercises in this tutorial: work from force vs. position graphs and calculating work from a force.
Want to join the conversation?
- Hi! The work = change in energy equation here is confusing me. Is that work energy theorem? Does work equal change in total energy, or only change in kinetic energy? I've also seen something about how work done by the non-conservative forces is change in energy? Are these all correct, and just reiterations of the same principle? What is the relationship between them? It's different everywhere I look and one source never seems to address the equation given in the other sources. Thanks!(6 votes)
- The Work Energy theorem states, work equals the change in kinetic energy. Also, from the Conservation of Energy equation, Ki+Ui+Wext=Kf+Uf, if you rearrange it, you get Wext=ΔE (total).(7 votes)
- why does negative sign means that it is giving energy to its surrounding? is the sign not for our ease only?..(i mean we assume that object moving toward left is negative and toward right is positive.)(1 vote)
- When work is negative, the force on the object is the opposite direction of the direction it is moving, so we can assume the object is slowing down. This kinetic energy must be going somewhere. So it is possible the energy is going to the object's surroundings, but it could also be going to potential energy.
Work is not a vector. It can be positive or negative the way temperature can be positive or negative. It is not an arbitrary convention.(6 votes)
- Why don't you, when you have a slanted force vector, break the vector down into its parallel and perpendicular parts, and then use the part that is parallel with the direction of motion in your work equation, instead of using the magnitude of the force?(3 votes)
- In the Common Mistakes section, should it be cos90 = 0 ?(2 votes)
- Yes you can write as cos 90=0 by rounding. But it is preferably used as -1. The exact value is equal to -0.44807361612.(2 votes)
- If a body is diagonally in air we have to take k.e or p.e to find work done(2 votes)
- Why do joules not account for differentiating surface frictions?(1 vote)
- When do you find the area under the force vs distance graph to find work and when do you simply use the formula w=Fd?(1 vote)
- how do you know whether or not to do addition or subtraction and how do you graph work(1 vote)
- if the area is under the x-axis, so then it will be negative. we just add all the work to get net work, but since it is negative work ( which means the force is opposite with the direction ) , it will look like subtraction because W+(-W)=W-W(0 votes)
- I know how to use simple shapes to find the area under functions on position vs. time graphs, but how could we use calculus to find the work for a function that might have fluctuating values of work? I already know calculus, and that integrals would be needed here, but I'm interested to see an actual application of them. What would the bounds look like, and how would we go about solving it?(0 votes)