A force of (6x^2 - 4x + 3) text{ N} acts on a | vedclass

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(A) The work done by a variable force is given by the integral of the force with respect to displacement: $W = \int_{x_1}^{x_2} F(x) dx$.Given $F(x) =

Vedclass · Accepted Answer

$201$

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(A) The work done by a variable force is given by the integral of the force with respect to displacement: $W = \int_{x_1}^{x_2} F(x) dx$.Given $F(x) = (6x^2 - 4x + 3) 	ext{ N}$,$x_1 = 2 	ext{ m}$,and $x_2 = 5 	ext{ m}$.$W = \int_{2}^{5} (6x^2 - 4x + 3) dx$$W = [\frac{6x^3}{3} - \frac{4x^2}{2} + 3x]_{2}^{5}$$W = [2x^3 - 2x^2 + 3x]_{2}^{5}$Substituting the limits:$W = [2(5)^3 - 2(5)^2 + 3(5)] - [2(2)^3 - 2(2)^2 + 3(2)]$$W = [2(125) - 2(25) + 15] - [2(8) - 2(4) + 6]$$W = [250 - 50 + 15] - [16 - 8 + 6]$$W = 215 - 14 = 201 	ext{ J}$.

$A$ force of $(6x^2 - 4x + 3) \text{ N}$ acts on a body of mass $0.75 \text{ kg}$ and displaces it from $x = 2 \text{ m}$ to $x = 5 \text{ m}$. The work done by the force is (in $\text{ J}$)

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