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tatiyna
3 years ago
5

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. initial

ly 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.)
Physics
1 answer:
IrinaVladis [17]3 years ago
5 0

\displaystyle W =\lim_{n\to\infty}{\sum_{i=0}^{n}{(678.16 - 36.26\;x_i)\cdot\dfrac{16}{n}}}.

<h3>Explanation</h3>

The mass comes in three parts:

  • the mass of the rope,
  • the mass of the water in the leaky bucket, and
  • the mass of the bucket.

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

  • m_\text{rope} = 0.7\;(16 - x),

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

  • m_\text{water} = 3.0\;(16 - x).

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

  • 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.

Weight of the bucket at height x:

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

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

  • F(x) = W(x) = 678.16 - 36.26\;x.

Express the work required as a definite integral:

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

Rewrite the definite integral as a Riemann Sum:

  • \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}}}.
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