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I'm trying to solve this problem in a Calculus textbook.

My final answer is 541.6 Ib-ft

However, the solutions page says that the answer is 550 Ib-ft

Is my solution incorrect?

If you're wondering, the textbook is Calculus with Analytic Geometry by George Simmions 2ed Section 7.7 problem 5 page 249

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1 Answers 1

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Let $w$ be the weight of the body, then $$\frac{dw}{dx}=k$$ where $k$ is a constant. When the bucket starts it weight is $w(0)=5+60=65$ lb, then, in the top $\frac13(60)=20$ lb of sand has leaked, so the weight of the bucket is $w(10)=5+40=45$ lb. It follows $$\int_{65}^{45}dw=\int_0^{10}kdx\qquad\implies\qquad-20=10k\qquad\implies\qquad k=-2$$

The weight of the bucket is given by $w(x)=65-2x$. Thus, the work is given by $$W=\int_0^{10}(65-2x)dx=650-10^2=550\quad\text{lb-ft.}$$

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    Thanks But why isn't the rate of weight change dw / dx = -65/30 Ib per 1 ft and therefore w(x) = -65/30 x + 65 – 2017-02-25
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    Since only the amount of sand decreases. On the other hand, $$\frac{dw}{dt}=-\frac{65}{30}\;\;\text{lb/ft}$$ is wrong cause 65 is the weight of both the bucket and the sand. – 2017-02-25