The loaded cab of an elevator has a mass of ๐๐ = 3.50 โ 103๐๐ and moves ๐ท๐ = 215๐ up theshaft in ๐ก๐ = 25๐ at constant speed. At what average rate does the force from the cable do workon the cab? (2 sig. figs, scientific notation)
Question
The loaded cab of an elevator has a mass of ๐๐ = 3.50 โ 103๐๐ and moves ๐ท๐ = 215๐ up theshaft in ๐ก๐ = 25๐ at constant speed. At what average rate does the force from the cable do workon the cab? (2 sig. figs, scientific notation)
Solution
The work done by the force from the cable on the cab can be calculated using the formula for work, which is force times distance. However, we first need to find the force.
The force exerted by the cable to lift the cab is equal to the weight of the cab, which is mass times gravity.
Given: Mass of the cab, Me = 3.50 * 10^3 kg Distance, De = 215 m Time, te = 25 s Acceleration due to gravity, g = 9.81 m/s^2
Step 1: Calculate the force Force = Me * g Force = 3.50 * 10^3 kg * 9.81 m/s^2 = 3.4335 * 10^4 N
Step 2: Calculate the work done Work = Force * Distance Work = 3.4335 * 10^4 N * 215 m = 7.382025 * 10^6 J
Step 3: Calculate the average rate of work done The average rate of work done is the total work done divided by the time taken.
Average rate = Work / time Average rate = 7.382025 * 10^6 J / 25 s = 2.95281 * 10^5 J/s
So, the average rate at which the force from the cable does work on the cab is approximately 2.95 * 10^5 J/s (to 2 significant figures in scientific notation).
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