A 2-kg toy car accelerates from 0 to 5 m/s. How much work is done?
Accepted Solution
A:
The work is the difference between the final and initial kinetic energy:[tex]W=\Delta K = \dfrac{1}{2}mv^2_f-\dfrac{1}{2}mv^2_i=\dfrac{1}{2}m(v^2_f-v^2_i)[/tex]Substitute your values to get[tex]W=\dfrac{1}{2}\cdot 2(25-0)=25[/tex]