91Ó°ÊÓ

Mechanical energy is the sum of kinetic energy (energy of motion) and potential energy (energy stored in a system due to the location of its elements).

The change in the mechanical strength of the system equals the work done in the system by external force and gravitational force.

Here by step 1, the total energy is constant.

Also the potential energy is\(U\left( 0 \right) = 0\). Therefore,

\(\begin{aligned}{}E &= U\left( 0 \right) + K\left( 0 \right)\\ &= 0 + 0.14\\ = 14{\rm{ }}J\end{aligned}\)

Now, \(U\left( { - 1.50} \right)\) is given by

\(\begin{aligned}{}U\left( { - 1.50} \right) &= E - K\left( { - 1.50} \right)\\ &= 0.14 - 0\\ &= 0.14{\rm{ J}}\end{aligned}\)

Hence, potential energy at \(x = - 1.5{\rm{ }}m\) is \(U\left( { - 1.50} \right) = 0.14{\rm{ J}}\).

Graph of \({\rm{U}}\left( {\rm{x}} \right)\) is obtained for different values of \({\rm{x}}\) by subtraction of \({\rm{K}}\left( {\rm{x}} \right)\) and \({\rm{E}}\).

So, graph of \({\rm{U}}\left( {\rm{x}} \right)\) is given by,

The conservative force is equal to negative rate of change of potential energy.

\(F\left( x \right) = - \frac{{dU\left( x \right)}}{{dx}}\)

Now, derivative of a function is zero at maxima and minima.

From the graph obtained in step 3, the potential energy \(U\left( x \right)\) has two minima at \(x = 1\), \(x = - 1\) and one maxima at \(x = 0\).

So, force is zero at \(x = 0,1, - 1\).

Hence \(F = 0\) at \(x = 0,1, - 1\).

We know that conservative force is equal to negative rate of change of potential energy.

\(F\left( x \right) = - \frac{{dU\left( x \right)}}{{dx}}\)

Now, from above equation we know that \(F\left( x \right)\) is positive when \(U\left( x \right)\) is negative.

Now, from the graph obtained in step 3, the potential energy \(U\left( x \right)\) is increasing in\( - 1 < x < 0\) and \(1 < x < 1.5\). So, slope of \(U\left( x \right)\) is positive in these range. Therefore force is negative in these range.

Now, from the graph obtained in step 3, the potential energy \(U\left( x \right)\) is decreasing in\( - 1.5 < x < - 1\) and \({\rm{0}} < x < 1\). So, slope of \(U\left( x \right)\) is negative in these range. Therefore force is positive in these range.

Hence \(F\) is positive in \( - 1.5 < x < - 1\) and \({\rm{0}} < x < 1\) and negative in \( - 1 < x < 0\) and \(1 < x < 1.5\).

Total energy of the system is define by,

\(E = U\left( x \right) + K\left( x \right)\)

\(\begin{aligned}{}U\left( { - 1.30} \right) &= E - K\left( { - 1.30} \right)\\ &= 0.14 - 0.27\\ &= - 0.13{\rm{ }}J\end{aligned}\)

When total mechanical energy is conserved, then maximum kinetic or potential energy the sphere can reach is the same as the initial mechanical energy.

Here sphere is released at \(x = - 1.30{\rm{ }}m\).

So, initial potential energy is \(U\left( { - 1.30} \right) = - 0.13{\rm{ }}J\) which is at \(x = - 0.55{\rm{ }}m\).

Now, the point where potential energy is minimum is the point where kinetic energy is maximum.

At \(x = - 1\) and \(x = 1\) potential energy is minimum. And at \(x = - 1\) and \(x = 1\) kinetic energy is maximum.

So, maximum kinetic energy is \(0.39{\rm{ }}J\).