Answer:
The power transmitted to the pulley is 0.0455 hp.
Explanation:
Given;
mass attached to the rope, m = 5 lb
radius of the pulley, r = 3 in
constant rate of fall of the mass, v = 5 ft/s
acceleration due to gravity, g = 32.2 ft/s²
1 lbf = 32.2 lb.ft/s²
The power transmitted to the pulley is calculated as;
P = Fv
P = (mg)v
[tex]P = 5 \ lb \ \times \ 32.2 \ \frac{ft}{s^2} \ \times \ 5 \ \frac{ft}{s} \ \times \ \frac{1 \ lbf}{32.2 \ lb.ft/s^2} \ \ = 25 \ \frac{ft.lbf}{s} \\\\[/tex]
in horse power, the power transmitted is calculated as;
[tex]P = \frac{25 \ ft.lbf}{s} \ \times \ \frac{1 \ hp}{550 \ ft.lbf/s} \ \ = 0.0455 \ hp[/tex]
Therefore, the power transmitted to the pulley is 0.0455 hp.
