Determining Speed from Work and Energy

Problema e solução em inglês, como no livro original.

Find the speed of the package in Figure 7.4 at the end of the push, using work and energy concepts.

Resolva no papel primeiro. Depois abra a solução um passo de cada vez e pare assim que conseguir terminar sozinho.

Estratégia
Here the work-energy theorem can be used, because we have just calculated the net work, $W_{net}$ , and the initial kinetic energy, $\frac{1}{2} {m v_{0}}^{2}$ . These calculations allow us to find the final kinetic energy, $\frac{1}{2} {mv}^{2}$ , and thus the final speed $v$ .
Solução
The work-energy theorem in equation form is

$W_{net} = \frac{1}{2} {mv}^{2} - \frac{1}{2} {m v_{0}}^{2} .$
7.21

Solving for $\frac{1}{2} {mv}^{2}$ gives
$\frac{1}{2} {mv}^{2} = W_{net} + \frac{1}{2} {m v_{0}}^{2} .$
7.22

Thus,

$\frac{1}{2} {mv}^{2} = 92 . 0 J + 3.75 J= 95.75 J.$
7.23

Solving for the final speed as requested and entering known values gives

$\begin{array}{l} v & = & \sqrt{\frac{2 (95.75 J)}{m}} = \sqrt{\frac{191.5 kg \cdot m^{2} {/s}^{2}}{30.0 kg}} \\ = & 2.53 m/s. \end{array}$
7.24
Discussão
Using work and energy, we not only arrive at an answer, we see that the final kinetic energy is the sum of the initial kinetic energy and the net work done on the package. This means that the work indeed adds to the energy of the package.

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