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 in meter. Work done by this force to move the particle from y = 0 to y =1 m is

Moderate

A

25 J
B

20 J
C

30 J
D

5 J

Solution

Work done by varibale force is
$\large W = \int\limits_{{y_i}}^{{y_f}} {Fdy} \\\\ Here\;,{y_i} = 0,{y_f} = 1m \\\\ \therefore \,W\int\limits_0^1 {\left( {20 + 10y} \right)dy} = \left[ {20y + \frac{{10{y^2}}}{2}} \right]_0^1 = 25\,J$

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