91影视

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

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

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, $U=mgh$鈥�(颈)

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:

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

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

b.

Time between two turns $=2.0s$

So:

Total time required = Time between two turns $脳$Number of turns

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

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