Answer to Question #25633 in Mechanics | Relativity for markejerson
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.
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 %