A lift is pulled up to the top of a building by a constant force of 6400N at a constant speed of 2m/s. It takes 3mins to reach the top.
(a) how much work is done in raising the lift to the top of the building ?
(b) what is the power required ?
(a) Force * distance =
6400 N * (2 m/s*180 s)= 2.304*10^6 J
(b) Work/time = 2.304*10^6 J/180s
= ___ Watts
Thanks !!! :)
To find the answers to these questions, we'll use some basic formulas from physics. Here's how to calculate the work done and the power required:
(a) Work done (W) is given by the formula:
W = force × distance
In this scenario, the force is the constant force pulling the lift up, which is 6400N. The distance is the height of the building. However, we need to find the distance traveled by the lift first. We can do that by multiplying the speed by the time taken:
Distance = speed × time
In this case, the speed is given as 2m/s and the time taken is 3 minutes. However, we need to convert the time from minutes to seconds:
Time (in seconds) = 3 minutes × 60 seconds/minute = 180 seconds
Now we can calculate the distance traveled by the lift:
Distance = 2m/s × 180s = 360m
Substituting these values into the work formula:
W = 6400N × 360m = 2,304,000 joules (J)
Therefore, the work done in raising the lift to the top of the building is 2,304,000 joules.
(b) Power (P) is given by the formula:
P = work done / time
In this case, we have already calculated the work done as 2,304,000 joules, and the time taken is 180 seconds. Substituting these values:
P = 2,304,000J / 180s = 12,800 watts (W)
Therefore, the power required is 12,800 watts.