Answer :
Answer:
The value is [tex]W_N = 11849 \ J [/tex]
Explanation:
From the question we are told that
The mass of the external weight is [tex]m = 51 \ kg[/tex]
The height through which the mass drops is [tex]h = 5.6 \ m[/tex]
The number of revolution made is [tex]N = 10100 \ kg[/tex]
The diameter of the free moving piston is [tex]d = 0.51 \ m \ kg[/tex]
The distance moved by the free moving piston is [tex]s = 0.71 \ m \ kg[/tex]
The atmospheric pressure is [tex]P = 101 \ kPa = 101*10^{3}\ Pa [/tex]
Generally the workdone by the external weight is mathematically represented as
[tex]W_w =m * g * h[/tex]
= [tex] 51 * 9.8 * 5.6 [/tex]
= [tex]2799 N [/tex]
Generally the workdone by the free moving piston is mathematically represented as
[tex]W_p =P * A * s[/tex]
Here A is the cross-sectional area with value
[tex] A = \pi * \frac{d^2}{4}[/tex]
[tex] A = 3.142 * \frac{0.51^2}{4}[/tex]
So
[tex]W_p =101*10^{3} * 3.142 * \frac{0.51^2}{4} * 0.71[/tex]
=> [tex]W_p = 14651 [/tex]
So
The net workdone is mathematically evaluated as
[tex]W_N = -W_w+W_p [/tex]
The negative sign shows that it is acting in opposite direction to [tex]W_N[/tex]
So
[tex]W_N = -2799+14651 [/tex]
[tex]W_N = 11849 \ J [/tex]