Answer
Verified
64.8k+ views
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.
${U_m} = {U_k} + {U_p}$ (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.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
According to classical free electron theory A There class 11 physics JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main