Respuesta :
Answer:
(a). The average power of the elevator motor during this time interval is 5903.6 W.
(b). The power input from motor is [tex]1.115\times10^{4}\ W[/tex]
Explanation:
Given that,
Mass of elevator = 650 kg
Time = 3.00 sec
Speed of the elevator = 1.75 m/s
We need to calculate the acceleration of elevator
Using formula of acceleration
[tex]a = \dfrac{v_{f}-v_{i}}{t}[/tex]
[tex]a=\dfrac{1.75}{3.00}[/tex]
[tex]a=0.58\ m/s^2[/tex]
(a). We need to calculate the net force on elevator is
Using formula of force
[tex]T=mg+ma[/tex]
[tex]T=m(g+a)[/tex]
[tex]T=650(9.8+0.58)[/tex]
[tex]T=6747\ N[/tex]
We need to calculate average velocity
Using formula of average velocity
[tex]v'=\dfrac{v_{f}+v_{i}}{2}[/tex]
[tex]v'=\dfrac{1.75}{2}[/tex]
[tex]v'=0.875\ m/s[/tex]
We need to calculate the average power
Using formula of power
[tex]P=T\times v'[/tex]
[tex]P=6747\times0.875[/tex]
[tex]P=5903.6\ W[/tex]
(b). We need to calculate the power input from motor
Using formula of power
[tex]P=F\times v[/tex]
[tex]P=mg\times v[/tex]
[tex]P=650\times9.8\times1.75[/tex]
[tex]P=1.115\times10^{4}\ W[/tex]
Hence, (a). The average power of the elevator motor during this time interval is 5903.6 W.
(b). The power input from motor is [tex]1.115\times10^{4}\ W[/tex]
