- #1
iwonde
- 30
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Your job is to lift 30-kg crates a vertical distance of 0.90 m from the ground onto the bed of a truck. How many crates would you have to load onto the truck in one minute for the average power output you use to lift the crates to equal 0.50 hp?
The equation that I used is: P_av = (delta W)/ (delta t)
Since 1hp = 746W --> 0.50hp = 373 W
delta W = (P_av)(delta t) = 373 W x 60 s = 22380 J
Work required to lift one crate = mgh = 30kg x 9.8 m/s^2 x 0.9m = 264.6J
Number of crates = 22380J / 264.6 J = 85
85 is not the correct answer. What did I do wrong?
Thanks.
The equation that I used is: P_av = (delta W)/ (delta t)
Since 1hp = 746W --> 0.50hp = 373 W
delta W = (P_av)(delta t) = 373 W x 60 s = 22380 J
Work required to lift one crate = mgh = 30kg x 9.8 m/s^2 x 0.9m = 264.6J
Number of crates = 22380J / 264.6 J = 85
85 is not the correct answer. What did I do wrong?
Thanks.