Answer:
The net work done in the 2-kg object is [tex]25J[/tex]
Explanation:
Hi
Knowns
Mass, [tex]2kg[/tex], initial speed [tex]v_{i}=0m/s[/tex] and final speed [tex]v_{f}=5m/s[/tex]
By using the formula of kinetic energy, [tex]K=\frac{1}{2}mv^{2}[/tex], we can obtain that [tex]K_{i}=\frac{1}{2}mv_{i}^{2}=\frac{1}{2}(2kg)(0m/s)^{2}=0[/tex], and [tex]K_{f}=\frac{1}{2}mv_{f}^{2}=\frac{1}{2}(2kg)(5m/s)^{2}=25J[/tex].
As the net work by definition is, [tex]W_{net}=K_{f}-K_{i}=25J-0J=25J[/tex], is the net work done in the 2-kg object to speed it up from 0 m/s to 5 m/s
