A leaky 10-kg bucket is lifted from the ground to a height of 16 m at a constant speed with a rope that weighs 0.7 kg/m. initially the bucket contains 48 kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 16-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.)

Both the mass of the rope [tex]m_\text{rope}[/tex] and the mass of the water in the bucket [tex]m_\text{water}[/tex] varies with the height [tex]x[/tex] of the bucket. Express the two masses as a function of [tex]x[/tex]:

[tex]m_\text{rope} = 0.7\;(16 - x)[/tex],

The water in the bucket behaves like yet another rope of density [tex]48 \;\text{kg}/ 16 \;\text{m}= 3.0\;\text{kg}\cdot\text{m}^{-1}[/tex]. The mass of the water left in the bucket at height [tex]x[/tex] will be

[tex]m_\text{water} = 3.0\;(16 - x)[/tex].

The mass of the bucket is [tex]10\;\text{kg}[/tex]. Combining the three:

[tex]m(x) = m_\text{rope}(x) + m_\text{water}(x) + m_\text{bucket} \\\phantom{m(x)} = 0.7\;(16-x) + 3.0\;(16 - x) + 10\\\phantom{m(x)} = 69.2 - 3.7\;x[/tex].

Weight of the bucket at height [tex]x[/tex]:

[tex]F_\text{weight}(x) = m(x)\cdot g = 9.8\;(69.2 - 3.7\;x) = 678.16 - 36.26\;x[/tex].

The bucket moves upward at a constant speed. As a result,

[tex]F(x) = W(x) = 678.16 - 36.26\;x[/tex].

Express the work required as a definite integral:

[tex]\displaystyle W = \int_{0}^{16}{F(x) \cdot dx} = \int_{0}^{16}{(678.16 - 36.26\;x)\cdot dx}[/tex].

Rewrite the definite integral as a Riemann Sum:

[tex]\displaystyle W = \int_{0}^{16}{(678.16 - 36.26\;x)\cdot dx} \\\phantom{W}= \lim_{n\to\infty}{\sum_{i=0}^{n}{(678.16 - 36.26\;x_i)\cdot\frac{16-0}{n}}}\\\phantom{W} = \lim_{n\to\infty}{\sum_{i=0}^{n}{(678.16 - 36.26\;x_i)\cdot\frac{16}{n}}}[/tex].