Question Video: Finding the Work from a Graph Representing Force with Displacement | Nagwa Question Video: Finding the Work from a Graph Representing Force with Displacement | Nagwa

# Question Video: Finding the Work from a Graph Representing Force with Displacement Mathematics • Third Year of Secondary School

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The figure shows the magnitude of a force acting on a body as it moves a distance π. Given that the force is measured in newtons, and the distance is measured in meters, determine the work done in joules by the force to move the body from π = 0 to π = 7 m.

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### Video Transcript

The figure shows the magnitude of a force acting on a body as it moves a distance π. Given that the force is measured in newtons and the distance is measured in meters, determine the work done in joules by the force to move the body from π equals zero to π equals seven meters.

To calculate the work done by the force on the body, which we can represent with a capital π, we can recall that work is equal to force multiplied by distance. Looking at the diagram, we see we have force listed on the vertical axis and distance on the horizontal axis.

Since weβre considering the motion of the body from π equals zero to π equals seven meters, when we calculate the work done, weβll be calculating the total area under this curve. That will be equal to the force applied times the displacement or distance traveled.

To calculate this total area under the curve, itβs helpful to break that area up into two pieces. One is a rectangle, where π ranges from zero to four meters. And the other part is a triangle, where π ranges from four to seven meters.

If we consider first the rectangle, we see that, over that rectangle, the force is constant at nine newtons. So we can write that the total work π is equal to nine newtons multiplied by the distance the body travels while itβs feeling that force, four meters plus some constant value weβll calculate in a moment.

That constant value π is the work done as the body moves from π equals four meters to π equals seven meters. Since the area under that section of the curve is a right triangle, we can recall that the area of a triangle is equal to one-half its base times its height, or in our case one-half three meters times nine meters the height of that triangle.

When we calculate this expression, we find that π is equal to 49.5 joules. Thatβs the work done on the body as it moves from π equals zero to seven meters.

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