#### To determine

**To find:**

The work done

#### Answer

W=650000 ft-lb

#### Explanation

**1) Concept:**

Approximate the required work by using the concept of Riemann sum. Then express the work as an integral and evaluate it.

**2) Given:**

i) Weight of the cable 2 lb/ft

ii) Length of the cable 500 ft

**3) Definition 3:**

W≈∑i=1nfxi* ∆x

**4) Calculation:**

Let’s place the origin at the top and x-axis pointing downward as in the figure.

Divide the cable into small parts with length ∆x

If xi* is a point in the ith such interval, then all points in the interval are lifted by approximately the same amount, namely, xi*

The cable weighs 2 pound per foot

Therefore, the weight of ith part is 2 lb/ft∆x ft=2·∆x lb

By using definition 3,

Thus, the work done on the ith part, in foot-pounds, is 2·∆x·xi*=2xi*∆x

The total work done by adding all these approximations and letting the number of parts become large (so ∆x→0 )

limn→∞∑i=1n2xi*∆x=∫05002xdx=x20500=250000 ft-lb

The work needed to lift the coal is 800lb·500ft=400,000 ft-lb

Thus, the total work required is 250000+400000=650000 ft-lb

**Conclusion:**

The total work done is 650000 ft-lb