Question

The potential energy U in joule of a particle of mass 1kg moving in x-y plane obeys the law $U = 3x + 4y$, where $\left( {x,y} \right)$ are the coordinates of the particle in metre. If the particle is at rest at $\left( {6,4} \right)$ at time t=0, then:
A. The particle has constant acceleration
B. The particle has zero acceleration
C. The speed of the particle when it crosses the y-axis is $10m/s$
D. Coordinates of the particle at t=1 sec are $\left( {4.5,2} \right)$

Answer 1

We know that work is equal to the product of force and displacement. By writing that in the integral form, we have –
$W = \int {F.dx}$
If we differentiate the above with respect to displacement x, we have –
$F = \dfrac{{dW}}{{dx}}$
The Work-energy theorem states that work done by the body is equal to decrease in the potential energy in the body.
$W = - U$
$\therefore F = - \dfrac{{dU}}{{dx}}$

Complete step-by-step answer:
Let us consider each of these statements one-by-one and check their correctness.

i)Statement-1: The particle has constant acceleration
Given the expression for potential energy, we need to calculate the acceleration.
From the proof above, we know that –
$F = - \dfrac{{dU}}{{dx}}$
Since, force is a vector quantity and U is dependant on x and y, we can rewrite the equation as –
$F = - \dfrac{{dU}}{{dx}}\hat i - \dfrac{{dU}}{{dy}}\hat j$
Substituting the equation of $U = 3x + 4y$ and differentiating, we get –
$F = - \dfrac{d}{{dx}}(3x + 4y)\hat i - \dfrac{{dU}}{{dy}}(3x + 4y)\hat j$
$F = - 3\hat i - 4\hat j$
We know that, $F = ma$ and since, mass is scalar, dividing the force by mass $m = 1kg$
$a = \dfrac{F}{m}$
$a = - 3\hat i - 4\hat j$
Here, we see that the acceleration vector is a constant value.
Hence, this statement is correct.

ii)Statement-2: The particle has zero acceleration
We calculated the value of acceleration, $a = - 3\hat i - 4\hat j$ from the previous step. This proves that the particle cannot have zero acceleration.
Hence, this statement is incorrect.

iii)Statement-3: The speed of the particle when it crosses the y-axis is $10m/s$
Here, we use the basic equations of speed to calculate the speed of the particle.
When it crosses the y-axis, this means the displacement is $\left( {6,0} \right)$
The equation for displacement, $s = ut + \dfrac{1}{2}a{t^2}$
Applying the formula for the displacement, we get –
$s = ut + \dfrac{1}{2}a{t^2} \\\ 0 - 6 = 0 \times t + \dfrac{1}{2} \times - 3 \times {t^2} \\\ Solving, \\\ \dfrac{{3{t^2}}}{2} = 6 \\\ 3{t^2} = 12 \\\ {t^2} = \dfrac{{12}}{3} = 4 \\\ t = \sqrt 4 = 2\sec \\\$
Now, we will use the value of time t, in the equation, $v = u + at$ to calculate the velocity.
${v_x} = u + at = 0 + \left( { - 3 \times 2} \right) = - 6m{s^{ - 1}}$
${v_y} = u + at = 0 + \left( { - 4 \times 2} \right) = - 8m{s^{ - 1}}$
Now that we have got the components of the velocity, the resultant velocity is equal to –
$V = \sqrt {{v_x}^2 + {v_y}^2}$
Solving the equation, we get –
$V = \sqrt {{{\left( { - 6} \right)}^2} + {{\left( { - 8} \right)}^2}} \\\ V = \sqrt {36 + 64} \\\ V = \sqrt {100} = 10m{s^{ - 1}} \\\$
Therefore, this statement is correct.

iv)Statement 4- Coordinates of the particle at t=1 sec is $\left( {4.5,2} \right)$
Using the equation, $s = ut + \dfrac{1}{2}a{t^2}$ for the 2 coordinates x and y from the position of rest, we get the time –
$x - 6 = 0 \times t + \dfrac{1}{2} \times - 3 \times {1^2} \\\ Solving, \\\ x = 6 - \dfrac{3}{2} \\\ x = 6 - 1.5 = 4.5m \\\$
$y - 4 = 0 \times t + \dfrac{1}{2} \times - 4 \times {1^2} \\\ Solving, \\\ y = 4 - \dfrac{4}{2} \\\ y = 4 - 2 = 2m \\\$
Hence, the coordinates are $\left( {4.5,2} \right)$
So, this statement is correct.

The correct statements are: A, C, and D.

Note: The work-energy theorem states that work done by the body is equal to the loss in the potential energy. So, this lost potential energy is compensated for the increase in kinetic energy. So, it can also be said that when there is work done by the body, there is an increase in the kinetic energy in the system.