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# Solve integral in Eng...

Ok..it\'s in Eng...so if you can help me out..that would be awesome!! Here comes the problem...

Water is being drawn from a well 100 ft deep, using a bucket that scoops up 100 lb of water. The bucket is pulled up at the rate of 2 ft/s, but it has a hole in the bottom through which water leaks out at a rate of 0.5 lb/s. Determine how much work is done in pulling the bucket to the top of the well. Neglect the weight of the bucket, the weight of the rope, and any work done in overcoming friction.

Hint: let y=0 correspond to the surface of the water in the well, so that y=100 at ground level, and consider the infinitesimal amount of work dW done to raise the bucket an infinitesimal distance dy, from y to y+ dy.

### 1 Answer

- 1 decade agoFavorite Answer
1ft=0.3048m，1lb=0.45359kg

假設水桶拉起高度為y，再拉起高度dy所做的功為dw

則由W=FS公式及y = v*t => t = y/v = y/2 (每秒2ft)

dw=F*dy => dw=(100-0.5t)g*dy => dw=(100-0.5*y/2)g*dy (g為重力常數)

W=∫(0積到100)(100-0.5*y/2)g*dy

=g∫(0積到100)(100-y/4) dy (g提到外面去)

=g*(100y-y^2/8 |( 0積到100)

=(10000-100^2/8)g

將ft換成m，lb換成kg

故算出來的功要在換算成焦耳

即W=(10000-100^2/8)*0.3048*0.45359*9.8=11855.3(J)

速解版

畫F-t圖求梯型面積即為所做的功

由於是F(重量)呈線性下降

最初是100ft拉起來後剩75ft

又花了100/2=50秒

所以W=[(100+75)*50/2]*0.3048*0.45359*9.8=11855.3(J)

有不懂再提出吧