To find the maximum set speed a person can go down the stairs, we can use the concept of conservation of energy. The potential energy gained by going down each step should be equal to the kinetic energy at the bottom of the stairs.
The potential energy gained by going down each step is given by:
\( \Delta PE = mgh \),
where \( m \) is the mass of the person, \( g \) is the acceleration due to gravity, and \( h \) is the height of each step.
The kinetic energy at the bottom of the stairs is given by:
\( KE = \frac{1}{2} mu^2 \),
where \( u \) is the maximum set speed.
Setting the potential energy gained equal to the kinetic energy, we have:
\( mgh = \frac{1}{2} mu^2 \).
Simplifying the equation, we get:
\( u^2 = 2gh \).
Substituting the given values, we have:
\( u^2 = 2 \times 10 \, \mathrm{m/s^2} \times 0.15 \, \mathrm{m} \).
Solving for \( u \), we find:
\( u = \sqrt{3} \, \mathrm{m/s} \).
Therefore, the maximum set speed a person can go down the stairs is approximately \( \sqrt{3} \, \mathrm{m/s} \).
To find the average power the person should develop when climbing the stairs at the same speed, we can use the work-energy theorem. The work done by the person is equal to the change in potential energy.
The work done by the person is given by:
\( W = \Delta PE = mgh \).
The time taken to climb the stairs is equal to the height of the stairs divided by the speed:
\( t = \frac{h}{u} \).
The average power is given by:
\( P = \frac{W}{t} \).
Substituting the values, we have:
\( P = \frac{mgh}{\frac{h}{u}} \).
Simplifying the equation, we get:
\( P = mu \).
Substituting the given values, we have:
\( P = 70 \, \mathrm{kg} \times \sqrt{3} \, \mathrm{m/s} \).
Calculating the value, we find:
\( P \approx 121.24 \, \mathrm{W} \).
Therefore, the average power the person should develop when climbing the stairs at the same speed is approximately 121.24 W. Glad to help