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The potential...

Question

The potential energy of a particle of mass 5 kg moving in xy-plane is given as U = (7x + 24y) joule, x and y being in metre. Initially at t = 0, the particle is at the origin (0, 0) moving with a velocity of $(8.6^i + 23.2^j) m s^{- 1}$ . Then

The velocity of the particle at t =4 s, is

5 m s^{- 1}

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The acceleration of the particle is

5 m s^{- 2}

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The direction of motion of the particle initially (at t = 0) is at right angles to the direction of acceleration

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The path of the particle is circle

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Solution

The correct option is B The acceleration of the particle is $5 m s^{- 2}$
$f_{x} = - \frac{d v}{d x} = - 7 N$
$f_{y} = - \frac{d v}{\partial y} = - 24 N$
$F = \sqrt{f_{x}^{2} + f_{y}^{2}} = 25 N$
Acceleration $= \frac{F}{m} = \frac{25}{5} = 5 m / s^{2}$

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The potential energy of a particle of mass 5 kg moving in xy-plane is given as U = (7x + 24y) joule, x and y being in metre. Initially at t = 0, the particle is at the origin (0, 0) moving with a velocity of (8.6^i+23.2^j)ms−1. Then

The correct option is B The acceleration of the particle is 5ms−2fx=−dvdx=−7Nfy=−dv∂y=−24NF=√f2x+f2y=25NAcceleration =Fm=255=5m/s2

The potential energy of a particle of mass 5 kg moving in xy-plane is given as U = (7x + 24y) joule, x and y being in metre. Initially at t = 0, the particle is at the origin (0, 0) moving with a velocity of $(8.6^i + 23.2^j) m s^{- 1}$ . Then

The correct option is B The acceleration of the particle is $5 m s^{- 2}$
$f_{x} = - \frac{d v}{d x} = - 7 N$
$f_{y} = - \frac{d v}{\partial y} = - 24 N$
$F = \sqrt{f_{x}^{2} + f_{y}^{2}} = 25 N$
Acceleration $= \frac{F}{m} = \frac{25}{5} = 5 m / s^{2}$