(B) The work done $W$ by a variable force $F$ is given by the integral of force with respect to displacement: $W = \int F \, dx$.Given the force $F =

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$\frac{aL^2}{2} + \frac{bL^3}{3}$

(B) The work done $W$ by a variable force $F$ is given by the integral of force with respect to displacement: $W = \int F \, dx$.Given the force $F = ax + bx^2$,we integrate from the unstretched position $(x = 0)$ to the stretched position $(x = L)$:$W = \int_{0}^{L} (ax + bx^2) \, dx$$W = \left[ \frac{ax^2}{2} + \frac{bx^3}{3} \right]_{0}^{L}$Substituting the limits:$W = \left( \frac{aL^2}{2} + \frac{bL^3}{3} \right) - (0 + 0)$$W = \frac{aL^2}{2} + \frac{bL^3}{3}$

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

Medium

When a rubber band is stretched by a distance $x$,it exerts a restoring force $F = ax + bx^2$,where $a$ and $b$ are constants. The work done in stretching the rubber band by a distance $L$ from its unstretched position is:

A
$\frac{1}{2}(aL^2 + bL^3)$
B
$\frac{aL^2}{2} + \frac{bL^3}{3}$
C
$\frac{aL^2}{3} + \frac{bL^3}{5}$
D
$\frac{aL^2}{7} + \frac{bL^3}{9}$

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Work, Energy, Power and Collision

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Work Done by Variable Force and Force-Displacement Graph

$A$ block of mass $2 \ kg$ is initially at rest on a horizontal frictionless surface. $A$ horizontal force $\overrightarrow{F} = (9 - x^2) \hat{i} \ N$ acts on it, starting from $x = 0$. The maximum kinetic energy of the block between $x = 0$ and $x = 3 \ m$ in joules is:

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The figure shows the frictional force versus displacement for a particle in motion. The loss of kinetic energy in travelling over $s = 0$ to $20\, m$ will be......$J$.

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What will be the area under the graph of variable force versus position?

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The potential energy $U(x)$ of a system is represented in the figure. The force $F(x)$ acting on the system will be represented by:

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When a rubber band is stretched by a distance $x$,it exerts a restoring force $F = ax + bx^2$,where $a$ and $b$ are constants. The work done in stretching the rubber band by a distance $L$ from its unstretched position is:

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$A$ block of mass $2 \ kg$ is initially at rest on a horizontal frictionless surface. $A$ horizontal force $\overrightarrow{F} = (9 - x^2) \hat{i} \ N$ acts on it, starting from $x = 0$. The maximum kinetic energy of the block between $x = 0$ and $x = 3 \ m$ in joules is:

The figure shows the frictional force versus displacement for a particle in motion. The loss of kinetic energy in travelling over $s = 0$ to $20\, m$ will be......$J$.

Power delivered to a mass is $P = (3t^2 - 2t + 1) \text{ W}$. The change in kinetic energy from $t = 2 \text{ s}$ to $t = 4 \text{ s}$ is: .................... $J$

What will be the area under the graph of variable force versus position?

The potential energy $U(x)$ of a system is represented in the figure. The force $F(x)$ acting on the system will be represented by:

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