farmer carrying leaky sack

[galactus]

Here is a fun little problem I ran across. Give it a go if you wish.

A farmer weighing 180 pounds carries a sack of grain weighing 30 lbs up a circular helical staircase around a silo of radius 25 feet.

As he climbs, grain leaks from the sack at a rate of 1 lb per 10 feet of ascent. How much work is done by the farmer in climbing through a

vertical distance of 50 feet in exactly 3 revolutions.

 

[wjm11]

A farmer weighing 180 pounds carries a sack of grain weighing 30 lbs up a circular helical staircase around a silo of radius 25 feet.

As he climbs, grain leaks from the sack at a rate of 1 lb per 10 feet of ascent. How much work is done by the farmer in climbing through a

vertical distance of 50 feet in exactly 3 revolutions.

The 3 revolutions are irrelevant to the work done. Only the change in height matters.

The farmer loses 5 lb (50 ft x 1 lb/10 ft) of grain in his 50 ft ascent, so 25 lb of grain make it to the top, along with the farmer himself.

Part 1) mgh = (180 + 25)(50) = 10,250 lb-ft

Since the rate of grain loss was constant, the average height the “lost” grain was raised was 25 ft, so

Part 2) mgh = (5)(25) = 125 lb-ft.

Total work = 10,375 lb-ft

 

[galactus]

Very good, wjm. Your method is so easy. I went and used all that calculus stuff and arrived at the same solution.