Solving Your Crate Loading Problem - Get the Right Answer!

  • Thread starter iwonde
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In summary, for the given task of lifting 30-kg crates a vertical distance of 0.90 m from the ground onto the bed of a truck, using an average power output of 0.50 hp, 85 crates would need to be loaded onto the truck in one minute. This calculation is based on the equation P_av = (delta W)/ (delta t), where delta W represents the work required to lift one crate (mgh) and delta t represents the time of one minute. However, there may be a slight discrepancy in the answer due to rounding.
  • #1
iwonde
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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.
 
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  • #2
I get 84.58, which rounds to 85, same as you.

What makes you think this answer is wrong?
 
  • #3


Firstly, I want to commend you for using the correct equation to solve this problem. However, there seems to be a mistake in your calculation of the work required to lift one crate. The correct calculation should be: mgh = (30kg)(9.8m/s^2)(0.9m) = 264.6 J. This means that the work required to lift one crate is 264.6 J, not 264.6 kJ as you have calculated. Therefore, the correct number of crates that can be loaded in one minute to equal 0.50 hp would be 22380 J / 264.6 J = 84.6 crates. This can be rounded up to 85 crates. I hope this clarifies your doubt.
 

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You can account for variations in crate weight and size by using flexible mathematical models that can adjust to different parameters. It may also be helpful to conduct sensitivity analyses to determine the impact of these variations on your loading solution and make adjustments accordingly.

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