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Question: The potential energy of mass 1 kg moving along the x- axis given by \(U(x) = \left[ {\dfrac{{{x^2}}}...

The potential energy of mass 1 kg moving along the x- axis given by U(x)=[x22x]JU(x) = \left[ {\dfrac{{{x^2}}}{2} - x} \right]J. If the total mechanical energy of a particle is 2 J, then find its maximum speed.

Explanation

Solution

Hint: - An object can store energy as a result of its position. This stored energy of position is referred to as potential energy. Potential energy is the stored energy of position possessed by an object.
Formula used: - Kinetic energy = 12mv2\dfrac{1}{2}m{v^2}.

Complete step-by-step solution -
As we know, total mechanical energy = Kinetic Energy + Potential Energy
For maximum speed, kinetic energy should be maximum and potential energy should be minimum. That is,
U(x)=[x22x]JU(x) = \left[ {\dfrac{{{x^2}}}{2} - x} \right]J should be minimum. For this, we have to differentiate the given value.
Therefore, differentiating both the sides.
dU(x)dx=d[x22x]dx\to \dfrac{{dU(x)}}{{dx}} = \dfrac{{d\left[ {\dfrac{{{x^2}}}{2} - x} \right]}}{{dx}}
dU(x)dx=(2x21)\to \dfrac{{dU(x)}}{{dx}} = \left( {\dfrac{{2x}}{2} - 1} \right)
dU(x)dx=(x1)\to \dfrac{{dU(x)}}{{dx}} = \left( {x - 1} \right)
For minimum potential energy, dU(x)dx=0\dfrac{{dU(x)}}{{dx}} = 0
Therefore, dU(x)dx=(x1)=0\dfrac{{dU(x)}}{{dx}} = \left( {x - 1} \right) = 0
(x1)=0\to \left( {x - 1} \right) = 0
x=1\to x = 1
Putting the value (x=1)\left( {x = 1} \right) in the equation U(x)=[x22x]JU(x) = \left[ {\dfrac{{{x^2}}}{2} - x} \right]J, we get
U(x)=[1221]JU(x) = \left[ {\dfrac{{{1^2}}}{2} - 1} \right]J
U(x)=[12]U(x) = \left[ {\dfrac{{ - 1}}{2}} \right]-----(i)
Now we know, total mechanical energy = Kinetic Energy + Potential Energy
Therefore, 2=K.E.+[12]K.E. + \left[ {\dfrac{{ - 1}}{2}} \right], {its is given in the question, total mechanical energy=2adn from (i)}
K.E.=2+[12]\to K.E. = 2 + \left[ {\dfrac{1}{2}} \right]
K.E.=[52]\to K.E. = \left[ {\dfrac{5}{2}} \right]----(ii)
Also, we know K.E.=12mv2K.E. = \dfrac{1}{2}m{v^2}----(iii)
From (ii) and (iii) we have
12mv2=52\dfrac{1}{2}m{v^2} = \dfrac{5}{2}
mv2=5\to m{v^2} = 5
1×v2=5\to 1 \times {v^2} = 5 (since in the question it is given that mass = 1 kg)
v2=5\to {v^2} = 5
v=5\to v = \sqrt 5

Therefore, the maximum speed of particles is 5ms\sqrt{5} \dfrac{m}{s}.

Note: - An object's kinetic energy is the energy it retains because of its motion, and potential energy is the energy an object maintains regardless of its location relative to other objects. Total mechanical energy, therefore = kinetic energy + potential energy. To answer these kinds of problems, we must note this basic principle.