Question

7. A man weighing 50 kg f supports a body of 25 kg f
on his head. What is the work done when he moves
a distance of 20 m up an incline of 1 in 10 ? Take
g = 9.8 ms-2
(Ans. 1470 )​

Answer 1

Answer:

1470 Joule

Step-by-step explanation:

Concept used: We need to calculate the vertical height ascended by the man to calculate the work

Total force due to weight = (50 + 25)g = 75g

The inclined plane slope is 1 in 20, if angle of inclination of the inclined plane is θ

Then

$\tan\theta=\frac{1}{10}$

$\theta=\tan^{-1}(\frac{1}{10})$

Therefore,

$\sin\theta=\frac{1}{\sqrt{101}}$

When the man moves 20 m up the inclined plane, if the vertical height of the man is h then

$\sin\theta=\frac{h}{20}$

or, $\frac{1}{\sqrt{101}}=\frac{h}{20}$

$\implies h=\frac{20}{\sqrt{101}}$

Work done by the man

$=75g\times h$

$=75\times 9.8\times \frac{20}{\sqrt{101}}$

$=75\times 9.8\times \frac{20}{\sqrt{101}}$

$=75\times 9.8\times \frac{20}{10.05}$

$=75\times 9.8\times \frac{20}{10.05}$

$=1470$   (approx)

thus work done by the man = 1470 Joule

Answer 2

Answer:

1470

Step-by-step explanation:

There is an incline of 1 in 10

Therefore, by moving a distance of 20m the person reaches a height of 2m

So, w=f*s=75*9.8*2 =1470