Question

The potential energy (in joules) of a particle of mass $1\,{\text{kg}}$ moving in a plane is given by $U = 3x + 4y$, the position coordinates of the point being x and y, measured in metres. If the particle is at rest at (6, 4), then, (This question has multiple correct options)
A. Its acceleration is of magnitude $5\,{\text{m/}}{{\text{s}}^2}$
B. Its speed when it crosses the y-axis is $10\,{\text{m/s}}$
C. Co-ordinates of the particle at $t = 1\,{\text{sec}}$ are (4.5, 2)
D. the particle has zero acceleration

Answer 1

Use the relation between the force and potential energy and determine net force and then acceleration of the particle. Determine the coordinates of the particle when it crosses the y-axis and displacement of the particle and then determine its final velocity using a kinematic equation. Determine the coordinates of the particle using the kinematic equation for displacement in x and y direction.

Formulae used:
The force $F$ acting on a particle is given by
$F = - \dfrac{{dU}}{{dr}}$ …… (1)
Here, $U$ is the potential energy of the particle and $r$ is the position of the particle.
The expression for Newton’s second law of motion is
${F_{net}} = ma$ …… (2)
Here, ${F_{net}}$ is the net force acting on a particle, $m$ is the mass of the particle and $a$ is acceleration of the particle.
The distance between the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is given by
$s = \sqrt {{{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{x_2} - {x_1}} \right)}^2}}$ …… (3)
The kinematic equation for final velocity $v$ is
${v^2} = {u^2} + 2as$ …… (4)
Here, $u$ is initial velocity, $a$ is acceleration and $s$ is displacement of the particle.
The kinematic equation for displacement $s$ is
$s = ut + \dfrac{1}{2}a{t^2}$ …… (5)
Here, $u$ is initial velocity, $a$ is acceleration and $t$ is time.

Complete step by step answer:
We have given that the potential energy of the particle is
$U = 3x + 4y$
The mass of the particle is
$m = 1\,{\text{kg}}$
The x-coordinate of force ${F_x}$ acting on the particle is given by
${F_x} = - \dfrac{{dU}}{{dx}}$
Substitute $3x + 4y$ for $U$ in the above equation.
${F_x} = - \dfrac{{d\left( {3x + 4y} \right)}}{{dx}}$
$\Rightarrow {F_x} = - 3\,{\text{N}}$

The y-coordinate of force ${F_y}$ acting on the particle is given by
${F_y} = - \dfrac{{dU}}{{dy}}$
Substitute $3x + 4y$ for $U$ in the above equation.
${F_y} = - \dfrac{{d\left( {3x + 4y} \right)}}{{dy}}$
$\Rightarrow {F_y} = - 4\,{\text{N}}$
The net force acting on the particle is given by
${F_{net}} = \sqrt {F_x^2 + F_y^2}$
Substitute $ma$ for ${F_{net}}$ in the above equation.
$ma = \sqrt {F_x^2 + F_y^2}$
$\Rightarrow a = \dfrac{{\sqrt {F_x^2 + F_y^2} }}{m}$

Substitute $- 3\,{\text{N}}$ for ${F_x}$, $- 4\,{\text{N}}$ for ${F_y}$ and $- 3\,{\text{N}}$ for $m$ in the above equation.
$\Rightarrow a = \dfrac{{\sqrt {{{\left( { - 3\,{\text{N}}} \right)}^2} + {{\left( { - 4\,{\text{N}}} \right)}^2}} }}{{1\,{\text{kg}}}}$
$\Rightarrow a = \dfrac{{\sqrt {9 + 16} }}{{1\,{\text{kg}}}}$
$\Rightarrow a = 5\,{\text{m/}}{{\text{s}}^2}$
Therefore, the acceleration of the particle is of magnitude $5\,{\text{m/}}{{\text{s}}^2}$. Hence, the statement given in option A is correct.

When the particle crosses the y-axis, its coordinates become (0. -4) as the x-coordinates of the particle becomes zero. The distance between the points (6, 4) and (0, -4) is given by using equation (3).
Substitute -4 for ${y_2}$, 4 for ${y_1}$, 0 for ${x_2}$ and 6 for ${x_1}$ in equation (3).
$s = \sqrt {{{\left( { - 4 - 4} \right)}^2} + {{\left( {0 - 6} \right)}^2}}$
$\Rightarrow s = \sqrt {64 + 36}$
$\Rightarrow s = \sqrt {100}$
$\Rightarrow s = 10\,{\text{m}}$
Hence, displacement of the particle after crossing the y-axis is $10\,{\text{m}}$.

The initial velocity of the particle before it crosses the y-axis is zero as it is at rest position.
$u = 0\,{\text{m/s}}$
Substitute $0\,{\text{m/s}}$ for $u$, $5\,{\text{m/}}{{\text{s}}^2}$ for $a$ and $10\,{\text{m}}$ for $s$ in equation (4).
${v^2} = {\left( {0\,{\text{m/s}}} \right)^2} + 2\left( {5\,{\text{m/}}{{\text{s}}^2}} \right)\left( {10\,{\text{m}}} \right)$
$\Rightarrow v = \sqrt {100}$
$\Rightarrow v = 10\,{\text{m/s}}$
Therefore, the speed of the particle when it crosses the y-axis is $10\,{\text{m/s}}$.Hence, the statement given in option B is correct.

We can determine the new coordinates of a particle at a given time $t = 1\,{\text{sec}}$ using equation (5). Rewrite equation (5) for displacement in x-direction.
$x - {x_1} = ut + \dfrac{1}{2}{a_x}{t^2}$
Substitute $6$ for ${x_1}$, $0\,{\text{m/s}}$ for $u$, $1\,{\text{s}}$ for $t$ and $- 3\,{\text{m/}}{{\text{s}}^2}$ for ${a_x}$ in the above equation.
$x - 6 = \left( {0\,{\text{m/s}}} \right)\left( {1\,{\text{s}}} \right) + \dfrac{1}{2}\left( { - 3\,{\text{m/}}{{\text{s}}^2}} \right){\left( {1\,{\text{s}}} \right)^2}$
$\Rightarrow x = 6 - 1.5$
$\Rightarrow x = 4.5\,{\text{m}}$
Hence, the x-coordinate is 4.5.
Rewrite equation (5) for displacement in y-direction.
$y - {y_1} = ut + \dfrac{1}{2}{a_y}{t^2}$

Substitute $4$ for ${y_1}$, $0\,{\text{m/s}}$ for $u$, $1\,{\text{s}}$ for $t$ and $- 4\,{\text{m/}}{{\text{s}}^2}$ for ${a_y}$ in the above equation.
$y - 4 = \left( {0\,{\text{m/s}}} \right)\left( {1\,{\text{s}}} \right) + \dfrac{1}{2}\left( { - 4\,{\text{m/}}{{\text{s}}^2}} \right){\left( {1\,{\text{s}}} \right)^2}$
$\Rightarrow y = 4 - 2$
$\Rightarrow y = 2\,{\text{m}}$
Hence, the y-coordinate is 2. Therefore, the coordinates at time $t = 1\,{\text{sec}}$ are (4.5, 2). Hence, the statement given in option C is correct.

The net acceleration of the particle is $5\,{\text{m/}}{{\text{s}}^2}$ and not zero. Hence, the statement given in option D is incorrect.

Hence, the correct options are A, B and C.

Note: The students may think how we determined the coordinates of the particle when it crosses the y-axis as (0, -4). When the particle crosses the y-axis, its x-coordinates becomes zero and since the particle is in the first quadrant (from coordinates (6, 4)) and then enters the second quadrant, the sign of y-coordinate is negative.