Answer :
Answer:
The average power provided by the tension in the cable pulling the lift is = 714 W
Explanation:
Given data
Mass = 71 kg
Change in height = 123 m
When the lift moves in upward direction then in that case kinetic energy is constant & only potential energy changes.
Change in potential energy Δ PE = m g ( [tex]h_f - h _ i[/tex] )
Δ PE = 71 × 9.81 × 123
Δ PE = 85670.73 J
Time = 2 min = 120 sec
So average power is given by
[tex]P_{avg} = \frac{Change \ in \ potential \ energy}{Time}[/tex]
[tex]P_{avg} = \frac{85670.73}{120}[/tex]
[tex]P_{avg} = 714 \ W[/tex]
Therefore the average power provided by the tension in the cable pulling the lift is = 714 W
Answer:
The answer is 714.23
Explanation:
Power= Watts/Time
Watts=0.5mv(final)^2+mgh
m=mass=71
v(final)= the final velocity=?
g=gravity=9.8
h=height=123m
step 1. find the unknown velocity.
v(f)=change in distance/change in time
they started at 0for both of these values so...
123meters /120seconds=1.025 v
step 2. solve for power.
now we can find power (0.5mv(final)^2+mgh/Time)
[0.5(71)(1.025)^2+(71)(9.8)(123)]/120 seconds=714.2335599...