(C) The work done is given by the line integral $W = \int \overrightarrow F \cdot d\overrightarrow r = \int (-K(y\hat i + x\hat j)) \cdot (dx\hat i +

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(C) The work done is given by the line integral $W = \int \overrightarrow F \cdot d\overrightarrow r = \int (-K(y\hat i + x\hat j)) \cdot (dx\hat i + dy\hat j) = -K \int (y\,dx + x\,dy) = -K \int d(xy)$.Path $1$: From $(0, 0)$ to $(a, 0)$. Here $y = 0$,so $dy = 0$. The work done $W_1 = -K \int_0^a (0) dx = 0$.Path $2$: From $(a, 0)$ to $(a, a)$. Here $x = a$,so $dx = 0$. The work done $W_2 = -K \int_0^a (a) dy = -K[ay]_0^a = -Ka^2$.Total work done $W = W_1 + W_2 = 0 + (-Ka^2) = -Ka^2$.

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

Difficult

$A$ force $\overrightarrow F = - K(y\hat i + x\hat j)$ (where $K$ is a positive constant) acts on a particle moving in the $x-y$ plane. Starting from the origin,the particle is taken along the positive $x$-axis to the point $(a, 0)$ and then parallel to the $y$-axis to the point $(a, a)$. The total work done by the force $\overrightarrow F$ on the particle is

IIT 1998 All IIT

A
$ - 2Ka^2$
B
$2Ka^2$
C
$ - Ka^2$
D
$Ka^2$

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Work, Energy, Power and Collision

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Work Done by Variable Force and Force-Displacement Graph

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IIT 1998 Paper All IIT years →

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IIT 2019

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Medium

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DifficultIIT 2013

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DifficultJEE MAIN 2023

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Consider a force $F = Kx^3$, which acts on a particle at rest. The work done by the force for a displacement of $2 \,m$ is given that $K = 2 \,N \cdot m^{-3}$. (in $\,J$)

MediumAP EAMCET 2022

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$A$ particle is moved along a path $A-B-C-D-E-F-A$,as shown in the figure,in the presence of a force $\vec{F} = (\alpha y \hat{i} + 2 \alpha x \hat{j}) \ N$,where $x$ and $y$ are in meters and $\alpha = -1 \ N/m$. The work done on the particle by this force $\vec{F}$ will be . . . . . . Joule.

$A$ particle of mass $10\,g$ moves in a straight line with retardation $2x$,where $x$ is the displacement in $SI$ units. Its loss of kinetic energy for the above displacement is $\left(\frac{10}{x}\right)^{-n}\,J$. The value of $n$ will be $............$.

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