Question

A machine lifts a load of 200 kg in a vertically upward direction. If the machine is driven by a freely falling mass of 400 kg through a distance of 8 m in 4 s, Calculate:
(i) Force exerted by the falling mass.
(ii) Work done by the falling mass.
(iii) Power input of machine.
(iv) Power output of machine, if it is 75 % efficient. [Take g = 10 ms-2 ]​

Answer 1

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Answer 2

Given:

mass $m_{1}$=200kg

$m_{2}=400 \ kg$

distance= 8m

time= 4s

To Find:

Force, Work done, Power input, and Power output

Solution:

(i) Force exerted by falling mass

$=falling mass\times g=400\times9.8\\=3920 \ N$

(ii) Work done by the falling mass

$=force \times distance$

$=392\times8\\=3136\ J$

(iii)The power input of the machine

$=\dfrac{3200}{4}\\=800\ W$

(iv) The Power output of the machine at 75% efficiency

$=\dfrac{work\ done}{time}\times efficiency$

$=\dfrac{3136}{4} \times0.75$

$=588 \ W$.