First slide

Work done by Diff type of forces

Question

A force F = 20 + 10y acts on a particle in y-direction, where F is in newton and y is in metre. Work done by this force to move the particle from y = 0 to y = 1 m is [NEET 2019]

Moderate

A

5J
B

25J
C

20J
D

30J

Solution

Work done by a force F, which is variable in nature in
moving a particle from y₁ to y₂ is given by $W = \int_{y_{1}}^{y_{2}} F \cdot d y \dots (i)$
Given, force, $F = 20 + 10 y, y_{1} = 0 and y_{2} = 1 m$
Substituting the given values in Eq. (i), we get
$W = \int_{0}^{1} (20 + 10 y) d y = {[20 y + \frac{10 y^{2}}{2}]}_{0}^{1}$
$= 20 (1 - 0) + 5 (1 - 0)^{2} = 25]$
$∴$ Work done by the force will be 25 J.

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