91Ó°ÊÓ

The conservation of energy principle states that in a closed system, the energy of interacting objects or particles remains constant. Kinetic energy, sometimes known as the energy of motion, was the first type of energy to be identified.

Here we have \(U\left( x \right) = - \alpha {x^2} + \beta {x^3}\)where ,\(\alpha = 2\,{\rm{j/}}{{\rm{m}}^2}\) and \(\beta = 2\,{\rm{j/}}{{\rm{m}}^2}\)

Now, conservation of energy is given by:

\(E = {K_i} + {U_i} = {K_f} + {U_f}\)

Where \(f\) stands for final and \(i\) stands for initial.

\({K_i},\,{K_f}\)is initial and final kinetic energy respectively.

\({U_i},\,{U_f}\)is initial and final potential energy respectively.

\(E\)is total energy,

The body was at rest initially at \(x = 0\), so \(E = 0\).

So, \({K_f} = - {U_f}\) and at point \(x = 4{\rm{ m}}\)

Now, we have \(U\left( x \right) = - \alpha {x^2} + \beta {x^3}\)

\(\begin{aligned}{}\frac{1}{2}m{v^2} &= - \left( { - \alpha {x^2} + \beta {x^3}} \right)\\v &= \sqrt {\frac{{2\left( {\alpha {x^2} - \beta {x^3}} \right)}}{m}} \\v &= \sqrt {\frac{{2\left( {2\,{\rm{j/}}{{\rm{m}}^2} \times {{\left( {4\,{\rm{m}}} \right)}^2} - 0.3\,{\rm{j/}}{{\rm{m}}^3} \times {{\left( {4\,{\rm{m}}} \right)}^3}} \right)}}{{0.09\,{\rm{kg}}}}} \\v &= 16.9\,{\rm{m/s}}\end{aligned}\)

Hence, the speed of object is \(16.9\,{\rm{m/s}}\).

The force on the body at point\(x = 4\,{\rm{m/s}}\)is given by

\(\begin{aligned}{}{F_x} &= {\left. { - \frac{{dU}}{{dx}}} \right|_{x = 4\,{\rm{m}}}}\\ &= - \frac{d}{{dx}}\left( {\alpha {x^2} + \beta {x^3}} \right)\\ &= - \left( { - 2\alpha x + 3\beta {x^2}} \right)\end{aligned}\)

Now, at \(x = 4\,{\rm{m/s}}\)

\(\begin{aligned}{}{F_x} &= - \left( { - 2\left( {2\,{\rm{j/}}{{\rm{m}}^2} \times \left( {4\,{\rm{m}}} \right)} \right) + 3\left( {0.3\,{\rm{j/}}{{\rm{m}}^3} \times {{\left( {4\,{\rm{m}}} \right)}^2}} \right)} \right)\\{F_x} &= 1.6\,{\rm{N}}\end{aligned}\)

Now, we know that

\(\begin{aligned}{}{F_x} &= m{a_x}\\{a_x} &= \frac{{{F_x}}}{m}\\{a_x} &= \frac{{1.6\,{\rm{N}}}}{{0.09\,{\rm{kg}}}} &= 17.8\,{\rm{m/}}{{\rm{s}}^{\rm{2}}}\end{aligned}\)

Also, here if object moves in +x-axis \(\frac{{dU}}{{dx}} < 0\)then \({F_x} > 0\). So, the object starts moving to the right side.

Hence the acceleration of object is \(17.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\) in the right side direction.

When the object is in motion, the maximum value of\(x\)at which object reach is called turning point.

At turning point object stops for some moment and start moving in opposite direction.

At this point kinetic energy of object is 0.

So, we have, \(E - U = 0\)

Also we have,

\(\begin{aligned}{}E = 0\\U = 0\end{aligned}\)

\(\begin{aligned}{} - \alpha {x^2} + \beta {x^3} &= 0\\x = 0{\rm{ and }}x &= \frac{\alpha }{\beta }\end{aligned}\)

Here we consider \(x = \frac{\alpha }{\beta }\)

\(\begin{aligned}{}x &= \frac{2}{{0.3}}\\ &= 6.67\,{\rm{m}}\end{aligned}\)

Hence the maximum value of\(x\)reached by the object during its motion is\(6.67\,{\rm{m}}\).