The Propelling force of a Rocket increases uniformly from zero to 5N in the first 12m and Remains Constant for the Next 40m . Find the total work done . Answer is 2300J . Plzzz Don't Answer for Points .
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Answered by
20
F ∝ x
F = kx
F = 5N at x = 12
So,
5 = 12x
x = 5/12
W₁ = ∫Fdx
= k∫x.dx
= (5/12) × [x²/2]₀¹²
= (5/12) × (12²/2)
= 30 Joule
W₂ = F × d
= 5 N × 40 m
= 200 Joule
Total work done = W₁ + W₂ = 230 Joule
[Note: I think an extra zero in your answer is typo]
F = kx
F = 5N at x = 12
So,
5 = 12x
x = 5/12
W₁ = ∫Fdx
= k∫x.dx
= (5/12) × [x²/2]₀¹²
= (5/12) × (12²/2)
= 30 Joule
W₂ = F × d
= 5 N × 40 m
= 200 Joule
Total work done = W₁ + W₂ = 230 Joule
[Note: I think an extra zero in your answer is typo]
Swapnil11111:
it's given in book
2300J
but 230 Joules is correct
maybe misprint in book
yeah
Integration dont come in ftre class 9
its correct swapnil !!
okk
Answered by
0
Answer:
ye the right answer is 230J
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