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,
[tex]W = F * d[/tex]
Where,
F= Force
d = Distance
The distance in this case is a composition between number of steps and the height. Then,
[tex]d=h*N[/tex], 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)
[tex]V_f^2-V_i^2 = 2a\Delta X[/tex]
Where,
[tex]V_f =[/tex] Final velocity
[tex]V_i =[/tex] Initial Velocity
a = Acceleration
[tex]\Delta X =[/tex] Displacement
PART A) For the particular case of work we know then that,
[tex]W = F*d[/tex]
[tex]W = m*g*(h*N)[/tex]
[tex]W = 50*9.8*(0.3*30)[/tex]
[tex]W = 4.41kJ[/tex]
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,
[tex]V_f^2-V_i^2 = 2a\Delta X[/tex]
Here,
[tex]V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow[/tex]3 steps in one second
[tex]v_f = 0[/tex]
Replacing,
[tex]V_f^2-V_i^2 = 2a\Delta X[/tex]
[tex]0-0.9^2=2a(30*0.3)[/tex]
Re-arrange for a,
[tex]a = -\frac{0.9^2}{2*30*0.3}[/tex]
[tex]a = -45*10^{-3}m/s^2[/tex]
At this point we can calculate the time, which is,
[tex]t = \frac{\Delta V}{a}[/tex]
[tex]t = \frac{0-0.9}{-45*10^{-3}}[/tex]
[tex]t = 20s[/tex]
With time and work we can finally calculate the power
P = \frac{W}{t} = \frac{4.41}{20}
P = 0.2205kW
