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
Answers
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}tanθ=
10
1
\theta=\tan^{-1}(\frac{1}{10})θ=tan
−1
(
10
1 )
Therefore,
\sin\theta=\frac{1}{\sqrt{101}}sinθ=
101
1
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}sinθ=
20
h
or, \frac{1}{\sqrt{101}}=\frac{h}{20}
101
1
=
20
h
\implies h=\frac{20}{\sqrt{101}}⟹h=
101
20
Work done by the man
=75g\times h=75g×h
=75\times 9.8\times \frac{20}{\sqrt{101}}=75×9.8×
101
20
=75\times 9.8\times \frac{20}{\sqrt{101}}=75×9.8×
101
20
=75\times 9.8\times \frac{20}{10.05}=75×9.8×
10.05
20
=75\times 9.8\times \frac{20}{10.05}=75×9.8×
10.05
20
=1470=1470 (approx)
thus work done by the man = 1470 Joule