Question

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

Options

• 25 J

• 20 J

• 30 J

• 5 J

Answer 1

25 J

Explanation:

Work done by variable force, W = `int_(y "initial")^(y "final")` Fdy

∴ W = `int_0^1 (20 + 10y)dy`

∴ W = `20 int_0^1 dy + 10 int_0^1 ydy`

∴ W = `20 [y]_0^1 + 10 ([y^2]_0^1)/2`

∴ W = `20 xx 1 + 10 xx 1/2`

∴ 25 J