Question

A ship of mass3×10000000 kg initially at rest is pulled by a force of 5×10000N through a distance of 3m . Assuring the resistance due to water to be negligible ,is the speed of the ship?

Answer 1

Given :-

» FORCE = F = 5×10^4 N

» MASS = M = 3×10^7 KG

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» WE KNOW THAT :-

=> F = Ma

=> 5×10^4 = 3×10^7×a

$= > a = \frac{5 \times {10}^{4} }{3 \times {10}^{7} } = 1.66 \times {10}^{ - 3}$
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» Distance travelled is = S = 3 m

» We know that :-

=> v² - u² = 2as

=> v² - 0 = 2×1.66×10^-3 × 3

=> v² = 9.96×10^-3

=> v = √9.96×10^-3 m/s
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Answer 2

Heya!
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⭐Given that ,

➡Mass Of Ship => 3 × 10^7 Kg .

➡Force => 5× 10 ⁴ N

➡Distance S => 3 m


⭐Using the Relation ,

F = ma

5 × 10⁴= 3 × 10^7 × a

$= > > a = \frac{5 \times 10 {}^{4} }{3 \times 10 {}^{7} }$


$= > > a = 1.66 \times 10 {}^{ - 3}$

Now we need to calculate the speed of ship , [ Final velocity ]

We have

Initial Velocity U = 0 [ As it starts from Rest ]

Distance s = 3m

Acceleration = 1.66 × 10^-3

Final velocity v = ?



➡Using the third equation of motion ,

V² - U² = 2as

=> 2 × 1.66 × 10^-3 × 3 = V²

=> 9.96 × 10^-3 = V²

$= > > v {}^{2} = 9.66 \times 10 {}^{ - 3} \\ \\ = > > v = \sqrt{9.66 \times 10 {}^{ - 3} }$
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