Question

An athlete needs to lose weight and decides to do it by “pumping iron.” (a) How many times must a $\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{kg}$weight be lifted a distance of $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$in order to burn off $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{lb}$of fat, assuming that that much fat is equivalent to $\mathbf{3500}\mathbf{}\mathbf{Cal}\mathbf{?}$(b) If the weight is lifted once every$\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{s}$, how long does the task take?

Short answer

a. $1.87\times {10}^4$times an 80.0 kg weight be lifted a distance of 1.00 m in order to burn 1.00 lb of fat.

b. If the weight is lifted once every $2.00\text{s}$, the total time required to complete the task will be 10.4 hours .

Step-by-step solution

Stating the given data

i. Mass of the weight, $m=80.0\text{kg}$.

ii. Lifting distance, $h=1.00\text{m}$.

iii. Amount of fat to be burned

$$\begin{gathered} 1lb=3500Cal \\ =3500Cal\times \frac{4186J}{1Cal} \\ =14.65\times {10}^{6}J \end{gathered}$$

iv. Time taken between two turns, $t=2.00\text{s}$.

Understanding the concept of energy

As the athlete will lift weight several times, it must be equal to the amount of fat to be burned. We convert the amount of fat to burn to its equivalent units and from this,we find the number of turns required.From the time gap between two turns and the number of turns calculated in the first part, we will get the total time required to complete the task.

Formula:

The potential energy of a body, $U=mgh$…(i)

Calculation of number of times of lifting the weight

a.

Assume the number of turns required is n.

So we can say that:

As the athlete will lift the weight several times, it must be equal to the amount of fat to be burned. So, the total energy to be energy using equation (i) is given and we can get the number of times of lifting as follows:

$$\begin{gathered} U=nmgh \\ 14.65\times {10}^6 \text{J}=n\times 80.0 \text{kg}\times 9.8 {\text{m/s}}^{\text{2}}\times 1.0 \text{m} \\ n=\frac{14.65\times {10}^6 \text{J}}{80.0\times 9.8\times 1.0 \text{J}} \\ =1.87\times {10}^4 \end{gathered}$$

Hence, the required number of times of lifting the weight is $1.87\times {10}^4$.

Calculation of total time required to complete the task

b.

Time between two turns $=2.0s$

So:

Total time required = Time between two turns $\times$Number of turns

$$\begin{gathered} \left(t\right)=2.00 \text{s}\times 1.87\times {10}^4 \\ =3.6\times {10}^4\text{s} \\ =3.74\times {10}^4\text{s}\times \frac{1 \text{hr}}{3600\text{s}} \\ =10.38\text{hr} \\ \approx 10.4\text{hrs} \end{gathered}$$

Hence, the required time to do the task is 10.4 hrs.