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.
​

Answer 1

Answer:

0.1m/s

Step 1: Acceleration of the ship, a

Applying Newton's law,

F=ma

⇒a=mF=3×1075×104=35×10−3m/s2

Step 2: Final speed of the ship

Using 3rd equation of motion,

v2−u2=2as

⇒v2−02=2×35×10−3m/s2×3m

⇒v2=10−2

⇒v=0.1m/s

Answer 2

Answer:

Given:-

• $\underline{ \sf \: M =3×10^{7} \: Kg}$

• S = 3m

• $\sf \: \: \: \: \: \: \: \: \: \: F = 5 × 10^{4}$

As initally it is at rest u = 0

----- First Find 'A'

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

We have,

• F = ma

$\\ \sf \longrightarrow \qquad 5 × 10^{4} =3×10^{7} × a$

$\\ \sf \longrightarrow \qquad a = \frac{5×10^{4}}{3×10^{7}}$

$\\ \sf \longrightarrow \qquad a = \frac{5}{3}× 10^{-3}$

---- Now Find 'v'

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

We have,

$\longrightarrow\qquad \boxed{ \underline{ \pink{ \sf \: \: v^{2}-u^{2}= 2as}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

As u = 0

$\\ \sf \longrightarrow \qquad \: v^{2}= 2as$

$\\ \sf \longrightarrow \qquad \: v^{2} = 2 × \frac{5}{3}× 10^{-3} × 3$

$\sf \longrightarrow \qquad v^{2} = 10 × 10^{-3}$

$\\ \sf \longrightarrow \qquad v^{2} = 10^{-2}\\ \\ \sf \longrightarrow \qquad v = \sqrt{10^{-2}} \\ \\\sf \longrightarrow \qquad \boxed{ \underline{ \purple{ \sf \: v = 10^{-1}}}}$

$\longrightarrow \qquad \underline {\boxed{\purple{ { \sf \: v = 0.1\: m/s}}}}$

Hence, The speed attained by the ship is V=0.1m/s