Question

A leaky 10-kg bucket is lifted from the ground to a height of 14 m at a constant speed with a rope that weighs 0.5 kg/m. Initially the bucket contains 42 kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 14-m level. Find the work done. (use 9.8 m/s2 for g.)

Answer 1

solution:

Weight of bucket = 10kg

Length or distance =14m

Weight of rope=0.5kg/m

At any point x of the rope,

=(0.5)(14-x)

=(7-0.5x)

Since the water finishes draining at 14m level and total weight of water is 42kg

Total mass=(7-0.5x)+(42-3x)+10=(59-3.5x)kg

Force=(9.8)(59-3.5x)

$work w =\lim_{n \to \infty }\sum_{i \to 1}^{n}(9.8)(59-3.5x)\Delta x\\=\int_{0}^{14}(9.8)(59-3.5x)dx\\=9.8\int_{0}^{14}(59-3.5x)dx\\9.8((59x-\frac{3.5x^2}{2})){_{0}}^{14}\\9.8(59(14)-\frac{3.5(14)^2}{2})\\=4733.4\\therefore,\\W=4733.4J\approx 4733J$