Answer:
The workdone is [tex]W_v = - 20 \ J[/tex]
Explanation:
From the question we are told that
The mass of the object is [tex]m = 2.5 \ kg[/tex]
The speed of fall is [tex]v = 2.5 \ m/s[/tex]
The depth of fall is [tex]d = 80\ cm = 0.8 \ m[/tex]
Generally according to the work energy theorem
[tex]W = \frac{1}{2} mv_2^2 - \frac{1}{2} mv_1^2[/tex]
Now here given the that the velocity is constant i.e [tex]v_1 = v_2 = v[/tex] then
We have that
[tex]W = \frac{1}{2} mv^2 - \frac{1}{2} mv^2 = 0 \ J[/tex]
So in terms of workdone by the potential energy of the object and that of the viscous liquid we have
[tex]W = W_v - W_p[/tex]
Where [tex]W_v[/tex] is workdone by viscous liquid
[tex]W_p[/tex] is the workdone by the object which is mathematically represented as
[tex]W_p = mgd[/tex]
So
[tex]0 = W_v + mgd[/tex]
=> [tex]W_v = - m * g * d[/tex]
substituting values
[tex]W_v = - (2.5 * 9.8 * 0.8)[/tex]
[tex]W_v = - 20 \ J[/tex]
