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Q.

A particle experiences a variable force $\vec{F} = (4 x \hat{i} + 3 y^{2} \hat{j})$ in a horizontal $x - y$ plane. Assume distance in meteres and force in newton. If the particle moves from the point $(1, 1)$ to point $(2, 2)$ in the $x - y$ plane, the kinetic energy changes by

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a

50 J

b

25.0 J

c

13.0 J

d

12.5 J

answer is D.

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Detailed Solution

By the work energy theorem
$Δ K = \int \bar{F} . d \bar{r}$
$Δ K = \int (4 x \hat{i} + 3 y^{2} \hat{j}) . (d x \hat{i} + d y \hat{j})$
$= \int_{1}^{2} 4 x d x + \int_{1}^{2} 3 y^{2} d y$
$= 4 {(\frac{x^{2}}{2})}_{1}^{2} + 3 {(\frac{y^{3}}{3})}_{1}^{2} = 2 (4 - 1) + (8 - 1) = 6 + 7 = 13 J$

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A particle experiences a variable force F→=(4xi^+3y2j^) in a horizontal x−y plane. Assume distance in meteres and force in newton. If the particle moves from the point (1,1) to point (2,2) in the x−y plane, the kinetic energy changes by