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Question 23

A force $$F = (5 + 3y^{2})$$ acts on a particle in the $$y$$-direction, where $$F$$ is in Newton and $$y$$ is in meter. The work done by the force during a displacement from $$y = 2$$ m to $$y = 5$$ m is ______ J.

Correct Answer: 132

Solution

$$W = \int_{y_i}^{y_f} F(y) \, dy$$

$$W = \int_{2}^{5} (5 + 3y^2) \, dy$$

$$W = \left[ 5y + \frac{3y^3}{3} \right]_2^5$$

$$W = \left[ 5y + y^3 \right]_2^5$$

$$W = 132 \text{ J}$$

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