Given data:
* The force applied by Kinley is F = 3000 N.
* The work done on the softball is W = 100 kJ.
Solution:
The work done on the softball in terms of the force and the displacement is,
[tex]W=F\times d[/tex]
where d is the displacement of the softball,
Substituting the known values,
[tex]\begin{gathered} 100\times10^3=3000\times d \\ d=\frac{100\times10^3}{3000} \\ d=\frac{100000}{3000} \\ d=\frac{100}{3} \end{gathered}[/tex]
By simplifying,
[tex]\begin{gathered} d=33.3\text{ m} \\ d\approx0.03\times10^3\text{ m} \\ d\approx0.03\text{ km} \end{gathered}[/tex]
Thus, the displacement of the ball is 33.3 meters or 0.03 km (in one significant figure).
