Acable that weighs 4 lbs/ft is used to lift a sandbag up a mine shaft 50 feet deep. a small tear in the bag causes sand to leak out at a constant rate as the sandbag is lifted. the sandbag weighs 80 pounds initially and 60 pounds when it gets to the top

W = 7500 lbs/ft

Explanation:

First, the question is incomplete. The missing part is How much work was done by lifting the bag this far?

Now that we need to calculate the work done to pull the sandbag, the best way to do this is assuming a linear function for the weight of the sandbag as it reach the height of the mine.

Now, the expression to calculate the work is:

W = W1(lift cable) + W2(lift sandbag)

To get work to lift the cable:

W = ∫2xdx

W1 = ∫2(50)dx = 50² = 2500 lbs/ft

To get the work to lift the bag, we should remember that we are going to assume that this is a linear function, so, y = ax + b

The work:

W = ∫F dx (1)

F = ax + b

b is the final weight of the bag, which is 60.

Solving for a:

a = m2 - m1 / x2 - x1

m1 and m2 are the mass of the sandbag, and x1 and x2 the distance of the sandbag.

We have this data already, so:

a = 80 - 60 / 50 - 0 = 0.4 or 2/5

F = 2/5x + 90 (2)

Replacing in (1)

W2 = ∫2/5x + 90 dx

W2 = ∫2/5x dx + ∫90 dx

W2 = 2/5 ∫x dx + 90 ∫dx

W2 = 2/5 x²/2 + 90∫dx

W2 = x²/5 + 90 x (from 0 to 50)

W2 = 50² / 5 + 90*50

W2 = 5000 lbs/ft

Finally the work done is:

W = W1 + W2

W = 2500 + 5000 = 7500 lbs/ft

are their multiple choices ?

explanation:

newton's first law of motion