Question

A well hole having a diameter of 5 cm is to be cut into the earth to a depth of 75 m. Determine the total work (in joules) required to raise the earth material to the surface if the average mass of 1 m is 1830 kg. (Data: g = 9.81 m/s) (Hint: How much work is required to raise a volume of ad/4 x dx from a depth of x feet to the surface?

Answer 1

Answer:

total work is 99.138 kJ

Explanation:

given data

diameter = 5 cm

depth = 75 m

density = 1830 kg/m³

to find out

the total work

solution

we know mass of volume is

volume = $\frac{\pi}{4} d^2 dx$

volume = $\frac{\pi}{4} d^2 1830 dx$

so

work required to rise the mass to the height of x m

dw = $\frac{\pi}{4} d^2 1830$ gx dx

so total work is integrate it with 0 to 75

w = $\int\limits^{75}_{0} {\frac{\pi}{4} d^2 1830 gx dx}$

w = $\frac{\pi}{4}$ × 0.05² × 1830× 9.81× $(\frac{x^2}{2})^{75}_0$

w = 99138.53 J

so total work is 99.138 kJ