A force $F = 20 + 10y$ acts on a particle in y-direction where F is in newton and y in meter. Work done by thi

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 $y = 1 	ext{ m}$ is

Accepted Answer

Work done by a variable force is $W = \int_{y_i}^{y_f} F dy$. Given $y_i = 0, y_f = 1 	ext{ m}$. $W = \int_{0}^{1} (20 + 10y) dy = [20y + \frac{10y^2}{2}]_{0}^{1} = 20(1) + 5(1)^2 = 25 	ext{ J}$.

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