A bag of sand originally weighing 144 lb is lifted at a constant rate. As it rises, sand also leaks out at a constant rate. At the instant when the bag has been lifted to a height of 18 feet, exactly half of the original amount of sand remains. (Neglect the weight of the bag and the lifting equipment.) How much work was done lifting the sand this far?

Step-by-step explanation:

The bag of sand wave is

W=144lb at x=0

The bag of sand is lifted at a constant rate, i.e the it is not accelerating, so a=0m/s²

Let W be the mass of the sand,

Sand leaks out at a constant rate

dW/dx= C

At a height=18ft, half of the original mass.

i.e x=18ft, M=72lb

We are asked to find the work at 18ft

Work is given as

Work=∫F•dx,

Where F is the force and also the weight of the sand, F=W

The Weight of the bag is a linear function,

dW/dx= C

Then, using variable separation

dW=Cdx

Integrating both sides

∫dW=∫Cdt

W=Cx + B

So, at x=0, W=144

144=B

B=144

W=Cx+144

Also at x=18, W=72lb

72=C×18+144

72-144=18C

-72=18C

C=-4

Then, the weight function becomes

W=-4x+144

W=144-4x

Then applying work formula

Work=∫W•dx

Work=∫(144-4x)dx. x=0 to x=18

Work = 144x-4x²/2. x=0 to x=18

Work=144x-2x² x=0 to x=18

Work=144(18)-2(18²) -0 -0

Work =2592-648

Work =1944 ft lbs

The work done in lifting the sand to 18ft is 1944 ft lbs

24/56=hx^7

step-by-step explanation: