Question

a crane can lift 10000 kg of coal in a1 hour from a mine of 180 m depth.if the efficiency of the crane is 80% its input power must be

Answer 1

potential energy=mgh
${10}^{4} \times 180 \times 10 = 18 \times {10}^{6}$
power =
$\frac{18 \times {10}^{6} }{60} = 3 \times {10}^{5}$
efficency= input power/total power x 100
$80 = \frac{x}{3 \times {10}^{5} } \times 100$
24x10^4

Answer 2

we know that,

efficiency of the crane = (Output power/Input power)*100%

80 = (output power/Input power)*100

Input Power = Output Power*(5/4)----- (1)

Now we have to calculate the value of 'Output Power'.

Output Power(P) = potential energy/time= mgh

Where m= mass of coal = 10000 kg

          g = acceleration due to gravity = 10 m/s²

          h = height = 180 m

              $P = mgh/t\\ P = 10000* 10*180/60\\P = 300000<strong> </strong>W$

Substituting 'P' value for equation (1),

   Input Power = 300000*(5/4) = 375000 W

Answer : Input Power = 375000 W