Question

A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without acceleration) a 100-kilogram weight 2-decimeters above the ground with an energy efficiency of 25%. How many repetitions can she do with the energy supplied from a single Oreo cookie? What happens to the number of repetitions that can be done if the efficiency increases?

Answer 1

Approximately
`325` (rounded down,) assuming that
`g = 9.81\; {\rm N \cdot kg^(-1)}`.

The number of repetitions would increase if efficiency increases.

Step-by-step explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules,
`{\rm J}`):

`\begin{aligned} & 53\; {\rm kCal} * \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} * \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 * 10^(5)\; {\rm J} \end{aligned}`.

Height of the weight (should be in meters,
`{\rm m}`):

`\begin{aligned} h &= 2\; {\rm dm} * \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}`.

Energy required to lift the weight by
`\Delta h = 0.2\; {\rm m}` without acceleration:

`\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} * 9.81\; {\rm N \cdot kg^(-1)} * 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}`.

At an efficiency of
`0.25`, the actual amount of energy required to raise this weight to that height would be:

.

Divide
`2.551 * 10^(5)\; {\rm J}` by
`784\; {\rm J}` to find the number of times this weight could be lifted up within that energy budget:

`\begin{aligned} \frac{2.551 * 10^(5)\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}`.

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.