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.)

the workdone =  4420.78 J

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 :

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 :

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

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

small = small current large mag large current