Answer to Question #25633 in Mechanics | Relativity for markejerson

Question

a hoisting engine lifts a 200 kg bucket containing 1200 Lof clay of density 2*10^3 kg m^3 to a height of 7 m in 20 sec. the power supplied to the engine is 12kW. calculate: work output in kJ, power output in kW and efficiency of the engine.

Answer 1

The work done by a constant force of magnitude F on a point that moves a distance d in the direction of the force is the product:
A = F*d
F = (m+M)g
m -& mass of& bucket
M - mass of clay
M = density*V,
V& - volume of clay
A = ( m+density*V)*g*d = (200 kg + 2100kg/m^3 *1.2m^3) 9.8m/s^2 7m = (200+2100*1.2)*9.8*7 J = 186592 J = 187 kJ
A = 187 kJ
P = F*v,
P - power, v - velocity.
v = d/t,
d - distance ( 7 m)
t - time (20sec)
P = F*d/t = A/t = 186592 J /20sec =& 9329.6 W = 9.33 kW
P = 9.33 kW
efficiency of the engine = P/P_in,
P_in - power supplied to the engine
efficiency of the engine = P/P_in = 9.33/12 = 0.7775 = 77.75 %
efficiency of the engine = 77.75 %