Answer:
8.2 *10^5 ft lb
Step-by-step explanation:
Wcoal = 800*500 = 400000 = 4*10^5 ft lb
[tex]Wrope =\int\limits^{400}_0 {6(400-y)} \, dy[/tex]
Using a right Riemann sum
The width of the entire region to be estimated = 400 - 0 = 400
Considering 8 equal subdivisions, then the width of each rectangular division is 400/8 = 50
F(x) = 6(400-y)
[tex]\left[\begin{array}{cccccccccc}x&0&50&100&150&200&250&300&350 &400\\f(x)&2400&2100&1800&1500&1200&900&600&300&0\end{array}\right][/tex]
Wrope = 50(2100) + 50(1800) + 50(1500) + 50(1200) + 50(900) + 50(600)
+ 50(300) + 50(0) = 420000 = 4.2 * 10^5 ft lb
Note: Riemann sum is an approximation, so may not give a accurate value
work done = Wcoal + Wrope= 4*10^5+ 4.2 * 10^5 = 8.2 *10^5 ft lb
The work done to lift the coal is 6.8*10^5 ft-lb
Data;
Let the distance (ft) below top of the shaft be represented by x
The weight of the coal to be lifted from mine = 500lb
The work done to lift the coal is
[tex]work = force * displacement\\w=fx\\force f(x) = 500 + 6x\\w_i= f(x_i)=[500+6x_i]\delta x\\[/tex]
Taking the summation limit
[tex]w = \lim_{0 \to 400} \sum [500+6x_i]\delta x\\ \delta w = f(x) \delta x\\[/tex]
we can take the integration of both sides where x will the range from 0 to 400.
[tex]\int \delta w = \int f(x) dx\\\int\limits^{400}_0{(500+6x)} \, dx \\w=[500x+3x^2]_0^4^0^0\\w=500(400-0)+ 3(400^2-0^2)\\w= 680000\\w=6.8*10^5 ft-lb[/tex]
From the calculations above, the work done is 6.8*10^5 ft-lb
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