91Ó°ÊÓ

We know the conservation of energy law given by,

\({K_1} + {U_1} = {K_2} + {U_2}\,\,\, \cdots \cdots \left( 1 \right)\) .

Where the kinetic energy is given by,

\(K = \frac{1}{2}m{v^2}\,\, \cdots \cdots \left( 2 \right)\)

The restoring force after putting the values of \(\alpha ,\,\,\beta \) , we get,

\(F\left( x \right) = - 60x - 18{x^2}\,\,\, \cdots \cdots \left( 3 \right)\)

We know that the conservative force is the negative derivative of the potential energy function.

Therefore,

\(\begin{aligned}{}F\left( x \right) = - \frac{{dU\left( x \right)}}{{dx}}\\ \Rightarrow dU\left( x \right) = - F\left( x \right)dx\end{aligned}\)

Integrating, we get,

\(U\left( x \right) = - \int {F\left( x \right)} dx + U\left( 0 \right)\)

Using \(\left( 3 \right)\) in the above equation, we get,

\(\begin{aligned}{}U\left( x \right) = - \int {\left( { - 60x - 18{x^2}\,} \right)\,} dx + U\left( 0 \right)\\U\left( x \right) = 30{x^2} + 6{x^3} + U\left( 0 \right)\end{aligned}\)

Now, using the condition \(U = 0\) at \(x = 0\)in above equation, we get,

\(U\left( x \right) = 30{x^2} + 6{x^3}\,\,\, \cdots \cdots \left( 4 \right)\)

Hence, the potential-energy function is, \(U\left( x \right) = 30{x^2} + 6{x^3}\,\).

We know the conservation of energy law given by,

\({K_1} + {U_1} = {K_2} + {U_2}\,\,\, \cdots \cdots \left( 1 \right)\) .

Where the kinetic energy is given by,

\(K = \frac{1}{2}m{v^2}\,\, \cdots \cdots \left( 2 \right)\)

The restoring force after putting the values of \(\alpha ,\,\,\beta \) , we get,

\(F\left( x \right) = - 60x - 18{x^2}\,\,\, \cdots \cdots \left( 3 \right)\)

We know that the conservative force is the negative derivative of the potential energy function.

Therefore,

\(\begin{aligned}{}F\left( x \right) = - \frac{{dU\left( x \right)}}{{dx}}\\ \Rightarrow dU\left( x \right) = - F\left( x \right)dx\end{aligned}\)

Integrating, we get,

\(U\left( x \right) = - \int {F\left( x \right)} dx + U\left( 0 \right)\)

Using \(\left( 3 \right)\) in the above equation, we get,

\(\begin{aligned}{}U\left( x \right) = - \int {\left( { - 60x - 18{x^2}\,} \right)\,} dx + U\left( 0 \right)\\U\left( x \right) = 30{x^2} + 6{x^3} + U\left( 0 \right)\end{aligned}\)

Now, using the condition \(U = 0\) at \(x = 0\)in above equation, we get,

\(U\left( x \right) = 30{x^2} + 6{x^3}\,\,\, \cdots \cdots \left( 4 \right)\)

Hence, the potential-energy function is, \(U\left( x \right) = 30{x^2} + 6{x^3}\,\).

Taking the initial position at point 1 and the position at which the speed is required at point 2.

Then we have,

\(\begin{aligned}{}{x_1} = 1.00,\,\,\,\,{v_1} = 0\\{x_2} = 0.50,\,\,\,\,{v_1} = ?\end{aligned}\)

Substituting the value of \({x_1}\) in \(\left( 4 \right)\), we get,

\({U_1} = 30{\left( 1 \right)^2} + 6{\left( 1 \right)^3}\,\,\, = 36J\).

Similarly, substituting the value of \({x_2}\) in \(\left( 4 \right)\), we get,

\({U_2} = 30{\left( {0.50} \right)^2} + 6{\left( {0.50} \right)^3}\,\,\, = 8.25J\).

Since the object is released from the rest, we get,

\({K_1} = 0\).

Now, substituting all the energy values in \(\left( 1 \right)\), we get,

\(\begin{aligned}{}0 + 36 = {K_2} + 8.25\\ \Rightarrow {K_2} = 27.75J\end{aligned}\)

Substituting the value of mass in \(\left( 2 \right)\), we get,

\(\begin{aligned}{l}27.75 = \frac{1}{2} \times \left( {0.90} \right) \times v_2^2\\ \Rightarrow {v_2} = \sqrt {\frac{{27.75 \times 2}}{{0.90}}} \\ \Rightarrow {v_2} = 7.85\end{aligned}\)

Hence, the speed of the object is, \({v_2} = 7.85\)m/s.