Question

a crane can lift up 10000kg of coal in 1 hour from a mine of 180 m depth. if the efficiency of the crane is 80%, it's input power must be : (take : g=10 ms-2)
a) 5kW
b) 6.25kW
c) 50kW
d) 62.5kW

Answer 1

Hey mate,

Potential energy = mgh 

Then,

10^4×10×180 = 18 × 10^6

Power = 18×10^6/60

= 3 × 10^5

Efficiency = Input Power/Total Power × 100

80 = x/3×10^5 ×100

x = 24x10^4

hope this helps you out!

Answer 2

Answer:

6.25kw

Step-by-step explanation:

Mass = 10000 kg

Depth = 180 m

g = 10

$Power = \frac{m \times g \times h}{t}$

$Power = \frac{10000 \times 10 \times180}{60 \times 60}$

Power = 5000w

We are given that only 80% of some power is used .

So, 0.8p = 5000

=> $p = \frac{5000}{0.8}$

p  = 6250w

p = 6.25kw

So, Option B is true .

Hence it's input power must be 6.25kw