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# The potential energy of a particle of mass 1Kg moving along x-axis is given by $U(x) = \left[ {\dfrac{{{x^2}}}{2} - x} \right]$. If the total mechanical energy of a particle is 20J then, find its maximum speed?

Last updated date: 13th Jun 2024
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Hint: Mechanical energy of a particle is given as the sum of the potential energy and kinetic energy.
${U_m} = {U_k} + {U_p}$ (Where ${U_m},{U_k},{U_p}$ are mechanical, kinetic and potential energies respectively)
Kinetic energy of the particle is given as: $\dfrac{1}{2}m{v^2}$
Using the above relation we will calculate the speed of the particle.

Complete step by step solution:
Mechanical energy is the sum total of the kinetic energy and potential energy of a particle. It is the energy of motion that determines how an object moves based on its motion and position. Mechanical energy comes into play when a force is acting on a body and then that force is transferred to kinetic and potential energy. When the object is at rest potential energy is stored in it and if the object is moving potential energy is converted to kinetic energy; overall energy of the system gets conserved.
Now, we will do the calculation part of the solution:
Mass of the particle is given as 1Kg.
Therefore, kinetic energy of the particle is given as:
$\Rightarrow {U_k} = {U_m} - {U_p}$(Mechanical energy minus potential energy)
We will substitute the values of each term.
$\Rightarrow {U_k} = \left[ {20 - (\dfrac{{{x^2}}}{2} - x)} \right]$.............(1)
To calculate the maximum speed we need to differentiate the kinetic energy and equate it to zero.
$\Rightarrow \dfrac{{d{U_k}}}{{dx}} = \dfrac{d}{{dx}}\left[ {20 - (\dfrac{{{x^2}}}{2} - x)} \right]$
$\Rightarrow \dfrac{{d{U_k}}}{{dx}} = x - 1 = 0$
$\Rightarrow x = 1$
On substituting the value of x in equation 1 to get the numerical value of kinetic energy:
$\Rightarrow {U_k} = \left[ {20 - (\dfrac{{{1^2}}}{2} - 1)} \right] \\ \Rightarrow {U_k} = \dfrac{{23}}{2} \\$(Value of x is substituted and obtained the numerical value).............(2)
Kinetic energy is given as:
$\dfrac{1}{2}m{v^2}$..........(3)
On equating equation 2 and 3
$\Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{{23}}{2}$
M is given as 1 so our value of speed is:
$\Rightarrow {v^2} = 23 \\ \Rightarrow v = \sqrt {23} m/s \\$
Maximum speed is $\sqrt {23}$ m/s.

Note: The best example of the combination of both kinetic and potential energy is: Hydropower Dam; where water is stored in the reservoir and then made to fall from height over the moving turbines. The stored water has kinetic energy in it and the falling water has kinetic energy in it which is then converted into electrical energy.