91Ó°ÊÓ

(a)

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

Hence, it can be calculated by integrating the force function and using the condition \(U\left( \infty \right) = 0\) in order to find the integration constant.

Therefore,

\(\begin{aligned}{}U\left( x \right) = - \int {F\left( x \right)} \,dx\\ = - \alpha \int {\frac{{dx}}{{{{\left( {x + {x_0}} \right)}^2}}}} \\ = \frac{\alpha }{{\left( {x + {x_0}} \right)}} + C\end{aligned}\)

where \(C\) is the integration constant.

By using \(U\left( \infty \right) = 0\), we get, \(C = 0\).

Therefore,

\(\begin{aligned}{}U\left( \infty \right) = 0 = \frac{\alpha }{{\infty + {x_0}}} + C\\ \Rightarrow C = 0\end{aligned}\)

Hence, \(U\left( x \right) = \frac{\alpha }{{x + {x_0}}}\).

(b)

Since we have the work done by the other forces is 0, then using the principle of mechanical energy conservation, which states that the change in kinetic energy must be equal to negative the change in the potential energy.

Thus,

\(\begin{aligned}{}\Delta K = - \Delta U\\\frac{1}{2}m\left( {v_f^2 - v_i^2} \right) = U\left( 0 \right) - U\left( {0.4} \right)\\\frac{1}{2} \times \left( {0.5} \right) \times \left( {v_f^2} \right) = \frac{{0.8}}{{0 + 0.2}} - \frac{{0.8}}{{0.4 + 0.2}}\\0.25v_f^2 = 2.67\\{v_f} = \sqrt {\frac{{2.67}}{{0.25}}} \,\,\\ = \,\,3.27\,{\rm{m/s}}\end{aligned}\)

The speed of the object is 3.27 m/s.