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

Answer:

Power, P = 228.67 Watts

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 :

$P=\dfrac{W}{t}$

$P=\dfrac{mgh}{t}$

$P=\dfrac{70\ kg\times 9.8\ m/s^2\times 4\ m}{12\ s}$

P = 228.67 watts

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