A 0.8-lbm object traveling at 200 ft/s enters a viscous liquid and is essentially brought to rest before it strikes the side. What is the increase in internal energy, taking the object and the liquid as the system?
Neglect the potential energy change.

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Accepted Answer

Conservation of energy requires that the sum of the kinetic energy and internal energy remain constant since we are neglecting the potential energy change. This allows us to write

[latex] E_{1}=E_{2} \quad \frac{1}{2} m V_{1}^{2}+U_{1}=\frac{1}{2} m V_{2}^{2}+U_{2} [/latex].

[latex] ext { The final velocity } V_{2} ext { is zero, so that the increase in internal energy }\left(U_{2}-U_{1} ight) ext { is given by } [/latex]

[latex] U_{2}-U_{1}=\frac{1}{2} m V_{1}^{2}=\left(\frac{1}{2} ight)(0.8 lbm )\left(200^{2} ft ^{2} / s ^{2} ight)=16.000 lbm - ft ^{2} / s ^{2} [/latex].

We can convert the above units to ft-lbf, the usual units on energy:

[latex] U_{2}-U_{1}=\frac{16,000 lbm - ft ^{2} / s ^{2}}{32.2 lbm - ft / s ^{2}- lbf }=497 ft - lbf [/latex].

Question 1.SP.15: A 0.8-lbm object traveling at 200 ft/s enters a viscous liqu...

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