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The potential energy v in joule of a particle of mass 1 kg moving in $x y$ plane, obeys the law $v = 3 x + 4 y$ , when $(x, y)$ are the co-ordinates of the particle in meter. If the particle is at rest at $(6, 4)$ at time $t = 0$ , then

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the co-ordinates of the particle at time $t = 4 s$ are $(- 18 m, - 28 m)$

the particle has constant acceleration

the work done by the external force from the position of rest of the particle and the instant at which the particle crossing x – axis is 25 joule

the speed of the particle when it crosses the y – axis is $10 m / s$

answer is A, B, C, D.

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Detailed Solution

a) Force acting on particle $\bar{F} = - (3 \hat{i} + 4 \hat{j}) N$

$F = constant \Rightarrow a = constant$

$b) a = \frac{F}{m} = \frac{- (3 i + 4 j)}{1}, | a | = \sqrt{9 + 18} = 5 m / s^{2}$

when particle crosses x -axis, $y = 0 \Rightarrow t_{1} = \sqrt{2} s$

$W = Fs = 5 \times 5 = 25$

c) Particle crosses y-axis $\Rightarrow x = 0 \Rightarrow t_{2} = 2 \sec$

$d) at t = 4 \sec, x = 6 - \frac{3}{2} (4)^{2} = - 18 m$

$y = 4 - 2 (4)^{2} = - 28 m$

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