A force $\to F$ = $k [k^i + y^j]$ , where k is a positive constant, acts on a particle moving in x-y plane starting from the point (3,5), the particle is taken along a straight line to (5, 7). The work done by the force is :

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A force F = −KX2(X≠0) acts on a particle in x-direction. The work done by this force in displacing the particle from x = +a to x = +2a is. (Where k is a positive constant)

The correct option is A −k2aGiven that , F=−Kx2 , Where x is the position of particle . Work done by this force , W=∫2aaF.dx=∫2aa−Kx2dx=[Kx]2aa=K2a−Ka=−K2a

A force F = $- \frac{K}{X^{2}} (X \neq 0)$ acts on a particle in x-direction. The work done by this force in displacing the particle from x = +a to x = +2a is. (Where k is a positive constant)

The correct option is A $\frac{- k}{2 a}$
Given that , $F = - \frac{K}{x^{2}}$ , Where $x$ is the position of particle .

Work done by this force , $W = \int_{a}^{2 a} F . d x = \int_{a}^{2 a} - \frac{K}{x^{2}} d x = [\frac{K}{x}]_{a}^{2 a} = \frac{K}{2 a} - \frac{K}{a} = - \frac{K}{2 a}$