‹ Class 9 · Ch 7
Work, Energy, and Simple Machines · Principle 3 of 33

Work is an area

The area under a force-displacement graph.

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NCERT: 7.1 Work Done by a Constant Force

Think

A graph of a push

A graph shows how large a force is at each position as an object moves along. The force is drawn on the up axis and the position on the sideways axis.

What do you think the area under this graph could tell us?

What this lesson covers

The idea

The work done by a force equals the area under its force–displacement graph between the initial and final positions, even when the force is not constant.

A graph of a push

A graph shows how large a force is at each position as an object moves along. The force is drawn on the up axis and the position on the sideways axis.

What do you think the area under this graph could tell us?

  • The work done
  • The mass of the object
  • The time taken

Shade the area

Choose the force and the displacement. Watch the shaded area.

Area is work

For a constant force the shaded area is a rectangle with height F and width s, so its area is F × s, which is the work. When the force changes, the area under the graph is still the work done between the first and last position.

The work done by a force is the area under its force-displacement graph between the initial and final positions, even when the force is not constant.

Notes

Work = the area under the force-displacement graph between the initial and final positions, even for a force that is not constant.

Check yourself

A constant force of 10 N acts over a displacement of 1 m. What is the area under the force-displacement graph, in J?

Answer: 10 J

Area = 10 N × 1 m = 10 J, the work done.

A constant force of 6 N acts over 4 m. What is the work done, in J?

Answer: 24 J

Area = 6 N × 4 m = 24 J.

The force on an object is not constant. How can we find the work done between two positions?

A force rises steadily from 0 N to 10 N over 4 m, so the area under its graph is a triangle. What is the work done, in J?

Answer: 20 J

Area of triangle = half × 4 m × 10 N = 20 J.

  • We cannot find it unless the force is constant. The book says the area under the graph still gives the work, even when the force is not constant.
  • Find the area under the force-displacement graph between them — correct. Yes!
  • Multiply the largest force by the final position. That is the area of a rectangle around the graph. For a force that rises from zero it is more than the area under the graph.
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