Answer to Question #198257 in Calculus for Ben

Question

Work 

Problem 1: A force of 500 dynes stretches a spring from its natural length of 20 cm to a length of 24 cm. Find the work done in stretching the spring from its natural length to a length of 28 cm.

Problem 2: A spring has a natural length of 6 in. A 1200-lb force compresses it to 5 1/2 in. Find the work done in compressing it from 6 in to 4 1/2 in.

Problem 3: An upright right-circular cylindrical tank of radius 5 ft and height 10 ft is filled with water. (a) How much work is done by pumping the water to the top of the tank? (b) Find the work required to pump the water to a level of 4 ft above the top of the tank.

Problem 4: A conical reservoir 10 m deep and 8m across the top is filled with water to a depth of 5 m. The reservoir is emptied by pumping the water over the top edge. How much work is done in the process?

Answer 1

Answer 1.)Given, that

$F = 500 \ dynes$ .

$500 = k(24-20)$

$\Rightarrow500 = 4k$

$\therefore k = \dfrac{500}{4} = 125$

Hence, $F(x) = 125x$

Therefore, total work done can be calculated as,

$W(x) = \int_{0}^{8}(125x)dx\\\ \\$

$W(x) = [\dfrac{125x^2}{2}]_0^8=\dfrac{125\times 8\times 8}{2}$

$Hence, \boxed{W = 4000}$

Answer 2.) We have,

$F = 1200$

We know that,

$F = kx$

$\Rightarrow 1200 = 5.5k$

$\Rightarrow k = \dfrac{1200}{5.5}\\\ \\$

$\therefore k = 218.18$

Hence, Total work done can be calculated as,

$W = \int_{6}^{4.5}218.18xdx$

$\Rightarrow W = 109.09[36-20.25]$

$\\Hence,\boxed{W = 1718.16}$

Answer3.)

a.)Work done can be calculated as:

$Height = 10$

Density of water = 62.4

$\Rightarrow \dfrac{5}{10} = \dfrac{x}{y}$

$\Rightarrow x = \dfrac{1}{2}y$

So, Work Done $= \int_{0}^{10} \pi x^2dy(62.4)(10-y)$

= $48.94 [333.33-2500]=40783.17$

Hence , $\boxed{Work\ done=40783.17}$

b.) $Radius = 5ft$

$Height = 10ft$

$\dfrac{5}{10} = \dfrac{x}{y}$

$x = \dfrac{1}{2}y$

Work done can be calculated as:

$= \int_{0}^{4}\pi x^2dy(62.4)(10-y)$

$= 48.94[213.33-64] = 7308.21$

4.) Given that,

Density of Water $= 1000\ kg/m^3$

$Radius = 4\ m$

$Height = 5\ m$

Hence,

$\Rightarrow\dfrac{4}{5} = \dfrac{x}{y}$

$\Rightarrow x = \dfrac{4}{5}y$

So, Work Done $= \int_{0}^{5}\pi x^2(1000)(5-y)dy$

$= \int_{0}^{5}\pi \cdot \dfrac{16}{25}\cdot y^2(1000)(5-y)dy$

$= 2009.6 \int_{0}^{5}(5y^2-y^3)dy$

$= 2009.6\times[\dfrac{5 \times(5)^3 }{3}-\dfrac{(5)^4}{4}]$

$= 104659.966J$

Hence, $\boxed{Work\ done=104659.966\ J}$