A force varying linearly with the distance acts on a body as shown above, The work done on the body by the force during the first 10m of motion is
The graph shows the force decreasing linearly from 20 N at 0 m to 0 N at 20 m.
For the first 10 m:
- Force at 0 m = 20 N
- Force at 10 m = 10 N (midway)
The area under the graph (work done) is the area of a trapezium:
\(\text{Work} = \frac{1}{2} \times (20 + 10) \times 10 = 15 \times 10 = \mathbf{150\, J}\).
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