Question

A leaky 10-kg bucket is lifted from the ground to a height of 13 m at a constant speed with a rope that weighs 0.8 kg/m. Initially the bucket contains 39 kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 13-m level. Find the work done. (Use 9.8 m/s2 for g.) Show how to approximate the required work by a Riemann sum. (Let x be the height in meters above the ground. Enter xi* as xi.)

Answer 1

the workdone = 4420.78 J

Step-by-step explanation:

The work done raising the rope.

Mass of the rope in the i-th subinterval is 0.8Δh and the work done raising this to the top is :

0.8Δh × 9.8 ×(13 -hi)

= 7.84 (13-hi) Δh

The total work done raising the rope is thus illustrated as :

`\lim_(n \to \infty) \sum ^n _( i=1)`

7.84 (13-hi) Δh

∫ 7.84 (13-hi) dh (from 0 to 13)

= -7.84 (1/2)(13 - h)² (from 0 to 13)

= 662.48 J

The mass of the water mi at height hi is given as mi = 39 - 3hi &

the workdone raising this through the distance Δh is 9.8(39 - 3hi)Δh

Thus, the total work done raising the water is :

`\lim_(n \to \infty) \sum ^n _( i=1)`

9.8(39 - 3hi)Δh

∫ 9.8 (39 - 3h) (from 0 to 13)

= -9.8 (1/6) (39 -3 h)² (from 0 to 13)

= 2484.30 J

The work done raising the bucket is 10 × 13 ×9.8 = 1274 J ,

The total work in raising the rope and the bucket of water is (1274 + 662.48 + 2484.30) J

= 4420.78 J

Thus, the workdone = 4420.78 J