Question

A trough is 7 meters long. 1.5 meters wide and 5 meters deep. The vertical cross-section of the trough parallel to an end is shaped like an isoceles triangle (with height 5 meters, and base, on top, of length 1.5 meters). The trough is full of water :). Find the amount of work in joules m3 required to empty the trough by pumping the water over the top. (Note: Use g = 9.8% as the acceleration due to gravity.) (density 1000 kg m Preview Joules

Answer 1

ANSWER :

A trough is 7 meter long. 1.5 meters wide. and 5 meters deep.

Take the slice of water of thickness

2

کرلی له کله WL

15 LI 커

$tan \frac{\theta }{2}=\frac{1.5/2}{5}=\frac{L/2}{x}$

$L=\frac{1.5/x}{5}\Rightarrow 0.3x$

Area of sllice = (L)(7) = 7L = 7(0.3.7) = 2.1.2

$Volume\, of\, sllice \Delta v= area.\Delta x\Rightarrow 2.1x\times \Delta x(m^{3})$

$Mass\, density \, of\, water \, \rho =1000kg/m^{3}$

$Mass\, \, of\, water \,slice \Delta m=\rho \Delta v\Rightarrow 2100x\Delta x$

$Weight\, \, of\, water \,slice \Delta w= \Delta m\times g\Rightarrow 2100g\Delta x$

Work done to pumpout the water slice (for pumpout slice to ) lifted "5-x" height.

$\Delta w= \Delta m\times height\Rightarrow\left [ 2100g\, x\Delta x \right ]\left [ 5-x \right ]$

$= 2100g\, (5x-x^{2})\Delta x$

Total work done for pumpout whole water

$W=\int_{0}^5{}2100 g(5x-x^{2})dx$

$W=2100 g\int_{0}^{5}(5x-x^{2})dx\Rightarrow 2100g\left ( \frac{5x^{2}}{2}-\frac{x^{3}}{3} \right )^{5}_{0}$

$W=2100 g \left ( \frac{5^{3}}{2}-\frac{5^{3}}{3} \right )\Rightarrow 2100g\left ( \frac{5^{3}}{6} \right )$

$W=700 g \left ( \frac{125}{2} \right )\Rightarrow 700(9.8)(62.5)joules$