Question

4. 9.222 m/s5. 5.523 m/s27. Answer: Dv²=85.04v² = 9.2217=19.22228.) Under the influence of some forces, the speedof a 21.325 kg object changed from 6 m/s to a certain speed.If the net work done on the object is 1450 J.Calculate its final speed.1. 13.115 m/s2. 15.634 m/s19.322 m/s4. 7.263 m/s3.5. 21.714 m/s29. Answer: A

Answer 1

By the work energy theorem we know that the net work done on an object is equal to its change in kinetic energy, that is:

$W=\Delta K$

where

$K=\frac{1}{2}mv^2$

Then, the work energy theorem can be express as:

$W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_0^2$

In this case we know that the work done is 1450 J, that the mass is 21.325 kg and the initial velocity is 6 m/s; plugging these values and solving for the final velocity we have:

$\begin{gathered} \frac{1}{2}(21.325)v_f^2-\frac{1}{2}(21.325)(6)^2=1450 \\ \frac{1}{2}(21.325)v_f^2=1450+\frac{1}{2}(21.325)(6)^2 \\ v_f^2=\frac{2}{21.325}(1450+\frac{1}{2}(21.325)(6)^2) \\ v_f=\sqrt{\frac{2}{21.325}(1450+\frac{1}{2}(21.325)(6)^2)} \\ v_f=13.115 \end{gathered}$

Therefore, the final velocity is 13.115 m/s and the correct option is A.