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(B) The work done by a variable force is given by the area under the force-displacement graph or the integral $W = \int F(x) dx$.For the first part $(

Vedclass · Accepted Answer

$5250$

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(B) The work done by a variable force is given by the area under the force-displacement graph or the integral $W = \int F(x) dx$.For the first part $(0 \leq x \leq 15\, m)$,the force is constant at $F = 200\, N$.$W_1 = 200 	imes 15 = 3000\, J$.For the second part $(15 The force function is $F(x) = 200 - \frac{200 - 100}{15}(x - 15) = 200 - \frac{100}{15}(x - 15) = 300 - \frac{100}{15}x$.$W_2 = \int_{15}^{30} (300 - \frac{100}{15}x) dx = [300x - \frac{100}{30}x^2]_{15}^{30}$.$W_2 = (300(30) - \frac{100}{30}(900)) - (300(15) - \frac{100}{30}(225)) = (9000 - 3000) - (4500 - 750) = 6000 - 3750 = 2250\, J$.The total work done is $W = W_1 + W_2 = 3000 + 2250 = 5250\, J$.

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