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Question
Solve the following problem.
While decreasing linearly from 5 N to 3 N, a force displaces an object from 3 m to 5 m. Calculate the work done by this force during this displacement.
Solution 1
For a variable force, work done is given by the area under the curve of force v/s displacement graph. From the given data, the graph can be plotted as follows:
∴ Work done, W = Area of `square`ABCD
∴ W = A(Δ AEB) + A `(square "EBCD")`
`= (1/2 xx "EB" xx "AE") + ("DE" xx "EB")`
`= (1/2 xx 2 xx 2) + (3 xx 2)`
= 8 J
Solution 2
Work done, W = Area of trapezium ADCB
∴ W = `1/2`(AD + CB) × DC
∴ W = `1/2`(5N + 3N) × (5m - 3m)
`= 1/2 xx 8 xx 2` = 8J
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