find the work needed to empty the trough

Question

A trough is 6 meters long, 3 meters wide, and 1 meters deep. The vertical cross-section of the trough parallel to an end is shaped like an isoceles triangle (with height 1 meters, and base, on top, of length 3 meters). The trough is full of water (density 1000 kg/m^3 ).

Find the amount of work in joules required to empty the trough by pumping the water over the top.

Use g = 9.8 m/s^2 as the acceleration due to gravity.

Answer 1

For each drop of water, the work is mass*gravity*lift, where lift is how far the drop is lifted. Let's use x to represent this lift (x=0 at the top of the trough and x=1 at the bottom of the trough). We imagine slicing the trough into (very) thin layers of thickness dx parallel to the ground. For each layer, the dimensions are 6 by (3-3x) by dx, and the lift is x, so:

dW = [1000*6*(3-3x)*dx]*9.8*x

dW = 1000*6*3*9.8*(1-x)*x*dx

W = 1000*6*3*9.8 ∫ x-x^2 * dx (from x=0 to x=1)

W = 18000*9.8*((1/2*[1]^2-1/3*[1]^3)-(1/2*[0]^2-1/3*[0]^3))

W = 18000*9.8*1/6

W = 3000*9.8

W = 29400 J

(We were lucky that all numbers were given in SI units)