To solve this exercise it is necessary to apply the equations concerning Work, both by general definition and by conservation of energy.
In other words, the work done by an object due to gravity is the equivalent to that defined by the potential energy equations, that is
[tex]W= mg\Delta h[/tex]
Where,
m=mass
g=gravitational acceleration
[tex]\Delta h =[/tex] Change in height
On the other hand we have that the work done by tension is defined by the conservation of kinetic and potential energy, that is to say
[tex]W= \Delta KE + \Delta PE[/tex]
Where,
[tex]\Delta KE =[/tex] Change in Kinetic Energy
[tex]\Delta PE =[/tex]Change in Potential Energy
PART A) As defined by the work done by gravity would be given by,
[tex]W = mg\Delta h[/tex]
[tex]W = (750)(9.8)(3.5)[/tex]
[tex]W = 25.725kJ[/tex]
Therefore the work done by gravity is 25.725kJ
PART B) The work done by the tension applies the energy conservation equation, that is to say
[tex]W= \Delta KE + \Delta PE[/tex]
[tex]W = (\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2)+(mgh_f-mgh_i)[/tex]
[tex]W = \frac{1}{2}m(v_f^2-v_i^2)+mg(h_f-h_i)[/tex]
Replacing with our values,
[tex]W = \frac{1}{2}(750)(0.75^2-0.25^2)+(750)(9.8)(3.5)[/tex]
[tex]W = 25.912kJ[/tex]
Therefore the work done by tension is 25.9kJ
(a) The amount of work done by gravity on the crane is 25,725 J.
(b) The amount of work done by tension is 25,912.5 J.
The amount of work done by gravity on the crane is calculated as follows;
W = mgh
W = 750 x 9.8 x 3.5
W = 25,725 J
The amount of work done by tension is calculated from conservation of mechanical energy as follows;
W = ΔK.E + ΔP.E
W = ¹/₂m(v² - u²) + mg(hf - hi)
W = ¹/₂ x 750 x (0.75² - 0.25²) + 750 x 9.8 x (3.5 - 0)
W = 187.5 + 25,725
W = 25,912.5 J
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