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

Hint: 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.