A removalist is loading five boxes onto a truck. Each has a mass of 10kg and a height of 30cm. The tray of the truck is 1.5m above the ground and the removalist is placing each box on top of the previous one.
What is the total work done by the removalist in lifting all the boxes onto the truck as described?
The answer i get is 441 J, but the textbook says 1029 J. Any help with this one is appreciated 
Thank you very much, both of you (but particularly Fred)
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Work = m*g*h (let’s use 10 for g…your book may use 9.8 m/s^2…we’ll soon see).
Here is a nice fast solution:
Each time we add a box, we need to lift it up 1.5 meters then 30 cm more for EACH new box. We can express this mathematically as…
WBox n (the work needed to lift up the n’th box)=
10*10*(1.5 +.3(n-1))= 150 + 30(n-1) = 120 + 30n
We want the sum of the first five of these…
5*120 + (30+60+90+120+150) = 1050 Joules
Yep, your book uses 9.8. OK? You get 5*117.6 + 29.4 + 58.8 + 88.2 +117.6 + 147 = 1029 J
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EXTRA CREDIT:
Now let’s be fancy (and also use 9.8)….The fancy part is invoking the equation for an aritmetic sequence so that IF they asked for 50 boxes instead of 5 we could get the answer WITHOUT having to do everyone seperately…which is a drag…’cause there are better things to do in life!
10*9.8*(1.5 +.3(n-1))= 117.6 + 29.4 n
Summing up a bunch of these is the same as multiplying 117.4 by whatever number of boxes you are loading and then adding on the sum of an aritmetic sequence.
To sum the first n terms of an aritmetic sequence (where d=29.4=the difference of adjacent terms), use the formula
Sn = 1/2 n (a1+an) where a1 = 1st term and an is the nth term…we want to sum 5 terms, so a1=29.4 and a5 = 147 = 5th term, thus S5 = 441
441 + 5 * 117.6 will be our answer which is 1029 Joules
which is what your book lists as the correct answer.
-Fred
EDIT: Do not include a whole bunch of decimal places like my friend below me. It is incorrect. (See "significant figures" in your textbook). In reality the answer (to two sig. figs. is 1030J) if you want to be a stickler.