Question

A rocket uses 400.0 J of chemical potential energy stored in the fuel while shooting the 0.55 kg rocket straight up

into the air. The rocket reaches a height of 23 m. What was the efficiency of the rocket in transforming the chemical
potential energy of the fuel into gravitational potential energy?
Select one:
O a. 25%
Ob. 35%
O c. 31%
O d. 29%

Answer 1

Approximately
`31\%` assuming that
`g = 9.81\; {\rm N \cdot kg^(-1)}`.

Step-by-step explanation:

Consider an object of mass
`m` in a uniform gravitational field of strength
`g`. If the height of that object increased by
`\Delta h`, the gravitational potential energy of that object would increase by
`m\, g\, \Delta h`.

In this question, the mass of the rocket is given to be
`m = 0.55\; {\rm kg}`. Assume that
`g = 9.81\; {\rm N \cdot kg^(-1)}`. The rocket has gained a height of
`\Delta h = 23\; {\rm m}`. Thus, the gravitational potential energy of this rocket would have increased by:

`\begin{aligned} m\, g\, \Delta h &= 0.55\; {\rm kg} * 9.81\; {\rm N \cdot kg^(-1)} * 23\; {\rm m} \\ &\approx 124.1\; {\rm J} \end{aligned}`.

In other words, the useful energy output from the combustion of the rocket fuel was approximately
`124.1\; {\rm J}`.

The energy input to this rocket was given to be
`400.0\; {\rm J}`. Thus, the efficiency of the energy conversion would be:

`\begin{aligned} \text{efficiency} &= \frac{(\text{useful energy out}\text{put})}{(\text{energy in}\text{put})} * 100\% \\ &\approx \frac{124.1\; {\rm J}}{400.0\; {\rm J}} * 100\% \\ &\approx 31\%\end{aligned}`.