To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.
In the case of work, we know that it is defined by,
![W = F * d](https://tex.z-dn.net/?f=W%20%3D%20F%20%2A%20d)
Where,
F= Force
d = Distance
The distance in this case is a composition between number of steps and the height. Then,
, for h as the height of each step and N number of steps.
On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)
![V_f^2-V_i^2 = 2a\Delta X](https://tex.z-dn.net/?f=V_f%5E2-V_i%5E2%20%3D%202a%5CDelta%20X)
Where,
Final velocity
Initial Velocity
a = Acceleration
Displacement
PART A) For the particular case of work we know then that,
![W = F*d](https://tex.z-dn.net/?f=W%20%3D%20F%2Ad)
![W = m*g*(h*N)](https://tex.z-dn.net/?f=W%20%3D%20m%2Ag%2A%28h%2AN%29)
![W = 50*9.8*(0.3*30)](https://tex.z-dn.net/?f=W%20%3D%2050%2A9.8%2A%280.3%2A30%29)
![W = 4.41kJ](https://tex.z-dn.net/?f=W%20%3D%204.41kJ)
Therefore the Work to do that activity is 4.41kJ
PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,
![V_f^2-V_i^2 = 2a\Delta X](https://tex.z-dn.net/?f=V_f%5E2-V_i%5E2%20%3D%202a%5CDelta%20X)
Here,
3 steps in one second
![v_f = 0](https://tex.z-dn.net/?f=v_f%20%3D%200)
Replacing,
![V_f^2-V_i^2 = 2a\Delta X](https://tex.z-dn.net/?f=V_f%5E2-V_i%5E2%20%3D%202a%5CDelta%20X)
![0-0.9^2=2a(30*0.3)](https://tex.z-dn.net/?f=0-0.9%5E2%3D2a%2830%2A0.3%29)
Re-arrange for a,
![a = -\frac{0.9^2}{2*30*0.3}](https://tex.z-dn.net/?f=a%20%3D%20-%5Cfrac%7B0.9%5E2%7D%7B2%2A30%2A0.3%7D)
![a = -45*10^{-3}m/s^2](https://tex.z-dn.net/?f=a%20%3D%20-45%2A10%5E%7B-3%7Dm%2Fs%5E2)
At this point we can calculate the time, which is,
![t = \frac{\Delta V}{a}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5CDelta%20V%7D%7Ba%7D)
![t = \frac{0-0.9}{-45*10^{-3}}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B0-0.9%7D%7B-45%2A10%5E%7B-3%7D%7D)
![t = 20s](https://tex.z-dn.net/?f=t%20%3D%2020s)
With time and work we can finally calculate the power
P = \frac{W}{t} = \frac{4.41}{20}
P = 0.2205kW