Answer:
D: 4.4m/s
Explanation:
From a known theorem the total work on a body is equal of the variation of kinetic energy:
[tex]W = \Delta K = K_f-K_i = \frac12 mv_f^2 - \frac12 m v_i^2[/tex]
Taking the first and last step in the chain of equalities and replacing the values we have, we get:
[tex]5 = \frac12(1)v_f^2 - \frac12(1)(3)^2\\10= v_f^2 -9 \rightarrow v_f^2 = 19 \rightarrow v_f=\sqrt19 \approx 4.4m/s[/tex]
