Question

How much power is required to lift a 70 kg bed to the top of a 4 meter flight of stairs in 12 seconds?

Answer 1

Given that:

mass (m) = 70 Kg ,

height (h) = 4 m = S ,

time (t) = 12 s,

Determine Power P = ?

P = rate of doing work = Work ÷ time

Work = F × S Joules

= (70 × 9.81 ) × 4 since F = m.g

= 2746.8 × 4

= 10987.2 J

Work = 10.98 KJ

Now, Power = 10.98 ÷ 12

P = 0.915 KW or 915 W

Answer 2

Power, P = 228.67 Watts

Step-by-step explanation:

It is given that,

Mass of bed, m = 70 kg

It is lifted to a top of 4 meter i.e. h = 4 m

Time taken, t = 12 seconds

We have to find the power required to lift the bed. The work done per unit time is called power. Mathematically, it is given by :

`P=(W)/(t)`

`P=(mgh)/(t)`

`P=(70\ kg* 9.8\ m/s^2* 4\ m)/(12\ s)`

P = 228.67 watts

Hence, the power required to lift the bed is 228.67 Watts.