Explanation:
It is given that,
Mas of the object, m = 6 kg
It is lifted through a distance, h = 5.25 m
Tension in the string, T = 80 N
(a) By considering the free body diagram of the object, the forces can be equated as :
[tex]T-mg=ma[/tex]
[tex]a=\dfrac{T-mg}{m}[/tex]
[tex]a=\dfrac{80-6\times 9.8}{6}[/tex]
[tex]a=3.33\ m/s^2[/tex]
Work done by tension, [tex]W_t=F\times h[/tex]
[tex]W_t=80\times 5.25[/tex]
[tex]W_t=420\ J[/tex]
(b) Work done by gravity, [tex]W_g=mgh[/tex]
[tex]W_g=6\times 9.8\times 5.25[/tex]
[tex]W_g=308.7\ J[/tex]
(c) Let v is the final speed of the object and u = 0
[tex]v=\sqrt{2ah}[/tex]
[tex]v=\sqrt{2\times 3.33\times 5.25}[/tex]
v = 5.91 m/s
Hence, this is the required solution.
