Question

A motor and gearbox system are used to raise a load from the ground to the roof. The load weighs 100Kg and the motor fed from supply of 110v and draws 15A. If the roof is 25m above the ground, calculate: a) The PE of the load when it has been raised, b) The amount of electrical energy used in that time

Answer 1

$a)$

$m= 100 kg$

$g= 9.8\frac{m}{s^2}$

$h = 25m$

$PE = mgh = 100*9.8*25= 24500J$

$\text{Answer: }PE = 24500J$

$b)$

$PE = 24500J$

$U= 110V$

$I = 15A$

$P=IU = 1650W=1.65kW$

$W= PE$

$W = Pt$

$t = \frac{W}{P}=\frac{24500}{1650}= 14.85s= 0.04125h$

$W = Pt= 1.65*0.04125=0.068\ kW*h$

$\text{Answer: }t = 14.55s; W= 0.068kW*h$