Answer:
total work is = 52450 J
Explanation:
given data
mass = 5000-lb
density = 10 lb/ft
height = 50 ft
solution
as we will treat here cable and ball are separate
and
here work need to lift cable is
w = (10Δy )(9.8 y ) j
and
now summing all segment of cable
so passing limit Δy to 0
so total work need
= [tex]\int\limits^{10}_0 {98y} \, dy[/tex]
= [tex][49 y^2]^{50}_0[/tex]
= 2450J
so lifting 5000 lb wrcking 50 m required additional 5000 + 2450
so total work is = 52450 J
The work done if the crane lifts the ball from ground level to 50 ft in the air by drawing in the cable is [tex]3.39 \times 10^5 \ J[/tex].
The given parameters;
The work done if the crane lifts the ball from ground level to 50 ft in the air by drawing in the cable is calculated as follows;
[tex]W = F\times d\\\\W = mg \times d[/tex]
where;
The work done is calculated as;
[tex]W = 5000 \times 32.17 \times 50\\\\W = 8,042,500 \ lb-ft^2/s^2\\\\W = \frac{0.13825 \ J/m}{1 \ lb.ft/s^2} \times 8,042,500 \ lb.ft^2/s^2\\\\W = 1,111,875.63 \ \frac{J-ft}{m} \times \frac{1 \ m}{3.28 \ ft} \\\\W = 338,986.47 \ J[/tex]
Thus, the work done if the crane lifts the ball from ground level to 50 ft in the air by drawing in the cable is [tex]3.39 \times 10^5 \ J[/tex].
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