Question

A conservative force \(\vec F\) is in the \( + x\)-direction and has magnitude \(F\left( x \right) = \frac{\alpha }{{{{\left( {x + {x_{0\,}}} \right)}^2}}}\), where \(\alpha = 0.800\)\(N \cdot {m^2}\) and \({x_{0\,}} = 0.200\)m. (a) What is the potential energy function \(U\left( x \right)\)for this force? Let \(U\left( x \right) \to 0\)as \(x \to \infty \). (b) An object with mass \(m = 0.500\)kg is released from rest at \(x = 0\) and moves in the \( + x\)-direction. If \(\vec F\) is the only force acting on the object, what is the object’s speed, when it reaches \({x_\,} = 0.400\)m?

Short answer

(a) The potential energy function is \(U\left( x \right) = \frac{\alpha }{{x + {x_0}}}\).

(b) The speed of the object is \({v_f} = 3.27\)m/s

Step-by-step solution

To find the potential energy function  

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

The speed of the object

(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.