Question

9. A ship of mass 3 x 107 kg and initially at rest can be
pulled through a distance of 3 m by means of a
force of 5 x 104 N. The water resistance is negligible.
Find the speed attained by the ship. ​

Answer 1

Correct Question :

A ship of mass 3 x 10⁷ kg and initially at rest can be pulled through a distance of 3 m by means of a force of 5 x 10⁴ N. The water resistance is negligible.

Find the speed attained by the ship.

Given :

• Mass of the ship = 3 × 10⁷ kg

• Distance = 3 m

• Force = 5 × 10⁴ N

To find :

Final Velocity/speed attained by the ship.

Solution :

First let us find the acceleration Produced by the ship.

We know that ,

$\boxed{\bf{F = ma}}$

Where :

• F = Force
• m = Mass
• a = Acceleration

By using the above equation and substituting the values in it, we get :

$:\implies \bf{F = ma}$

$:\implies \bf{5 \times 10^{4} = 3 \times 10^{7} \times a}$

$:\implies \bf{\dfrac{1}{a} = \dfrac{3 \times 10^{7}}{5 \times 10^{4}}}$

$:\implies \bf{\dfrac{1}{a} = \dfrac{3 \times 10^{7 - 4}}{5}}$

$:\implies \bf{\dfrac{1}{a} = \dfrac{3 \times 10^{3}}{5}}$

$:\implies \bf{a = \dfrac{5}{3} \times 10^{-3}}$

$\boxed{\therefore \bf{a = \dfrac{5}{3} \times 10^{-3}}}$

Hence, the acceleration Produced by the ship is 5/3 × 10-³ m/s²

To find the final Velocity of the ship.

We know the third Equation of Motion i.e,

$\boxed{\bf{v^{2} = u^{2} + 2aS}}$

By using the above equation and substituting the values in it, we get :

$:\implies \bf{v^{2} = u^{2} + 2aS}$

$:\implies \bf{v^{2} = 0^{2} + 2aS}$

[Note : Here the value of Initial Velocity is taken as 0 , since the the ship is starting from rest]

$:\implies \bf{v^{2} = 2aS}$

$:\implies \bf{v^{2} = 2 \times \dfrac{5}{3} \times 10^{-3} \times 3}$

$:\implies \bf{v^{2} = 6 \times \dfrac{5}{3} \times 10^{-3}}$

$:\implies \bf{v^{2} = \dfrac{30}{3} \times 10^{-3}}$

$:\implies \bf{v^{2} = 10 \times 10^{-3}}$

$:\implies \bf{v^{2} = 10 \times \dfrac{1}{1000}}$

$:\implies \bf{v^{2} = \dfrac{1}{100}}$

$:\implies \bf{v = \sqrt{\dfrac{1}{100}}}$

$:\implies \bf{v = \dfrac{1}{10}}$

$:\implies \bf{v = 0.1}$

$\boxed{\therefore \bf{v = 0.1\:ms^{-1}}}$

Hence, the final Velocity of the ship is 0.1 m/s.