(A) The work done by a variable force is given by the integral $W = \int_{x_1}^{x_2} F \, dx$.Given $F = (3x^2 + 2x - 5) \text{ N}$,$x_1 = 2 \text{ m}

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(A) The work done by a variable force is given by the integral $W = \int_{x_1}^{x_2} F \, dx$.Given $F = (3x^2 + 2x - 5) \text{ N}$,$x_1 = 2 \text{ m}$,and $x_2 = 4 \text{ m}$.$W = \int_{2}^{4} (3x^2 + 2x - 5) \, dx$Integrating the terms: $W = [x^3 + x^2 - 5x]_{2}^{4}$Evaluating at the upper limit $(x = 4)$: $(4)^3 + (4)^2 - 5(4) = 64 + 16 - 20 = 60$Evaluating at the lower limit $(x = 2)$: $(2)^3 + (2)^2 - 5(2) = 8 + 4 - 10 = 2$Subtracting the values: $W = 60 - 2 = 58 \text{ J}$.

Class 11 Physics Work, Energy, Power and Collision Work Done by Variable Force and Force-Displacement Graph

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$A$ force $(3x^2 + 2x - 5) \text{ N}$ displaces a body from $x = 2 \text{ m}$ to $x = 4 \text{ m}$. The work done by this force is .......... $J$.

JEE MAIN 2024 All JEE MAIN

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$58$
B
$59$
C
$60$
D
$61$

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Work Done by Variable Force and Force-Displacement Graph

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$A$ force $(3x^2 + 2x - 5) \text{ N}$ displaces a body from $x = 2 \text{ m}$ to $x = 4 \text{ m}$. The work done by this force is .......... $J$.

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The force-displacement graph for a particle is given. Calculate the work done by the force as the particle moves from $x = 1 \ cm$ to $x = 5 \ cm$. (in $erg$)

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