Chapter 15, Problem 51SGA

To determine

To Find: The kinetic energy of the object for the given period of time.

Answer to Problem 51SGA

  $1.25m\times {10}^{-5}$

Explanation of Solution

Given:

  $\begin{array}{l}k\left(t\right)=\frac{1}{2}m•v{\left(t\right)}^2\\ v\left(t\right):\text{Velocity of the object }\\ t:\text{time}\\ v\left(t\right)=\frac{50}{1+t^2}forallt\ge 0\\ t\to 100\end{array}$

Formula Used:

  $\underset{x\to a}{\lim }\left[f\left(x\right)\times g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)\times \underset{x\to a}{\lim }g\left(x\right)$

Calculation:

  $\begin{array}{l}k\left(t\right)=\frac{1}{2}m•v{\left(t\right)}^2\\ v\left(t\right):\text{Velocity of the object }\\ t:\text{time}\\ v\left(t\right)=\frac{50}{1+t^2}forallt\ge 0\\ t\to 100\\ \underset{t\to 100}{\lim }k\left(t\right)=\underset{t\to 100}{\lim }\left\{\frac{1}{2}m•{\left(\frac{50}{1+t^2}\right)}^2\right\}\\ =\frac{1}{2}m•{\left(\frac{50}{1+{100}^2}\right)}^2\approx \frac{1}{2}m•{\left(\frac{50}{10000}\right)}^2\\ =\frac{1}{2}m•\frac{25}{{10}^6}\\ =1.25m\times {10}^{-5}\\ \end{array}$

Conclusion:

Thus the kinetic energy at the required time is $1.25m\times {10}^{-5}$