First slide

Work done by Diff type of forces

Question

A force F=20+10 y acts on a particle in y -direction where F is in newton and y in meter. Work done by this force to move the particle from y=0 to y=1 m is

Moderate

A

20 J
B

30 J
C

5 J
D

25 J

Solution

$Given : F = 20 + 10 y work done, W = \int F \cdot dy = \int_{0}^{1} (20 + 10 y) dy = {[20 y + \frac{10}{2} y^{2}]}_{0}^{1} = 20 + \frac{10}{2} = 25 J$

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