Question

A ship of mass 3×10^7 kg initially at rest is pulled by a Force of 5×10^4 N through a distance of 3 m.Neglecting friction, the speed of ship at this moment is?

Answer 1

here 

m = 3 x 10⁷ kg 

F = 5 x 10⁴ N 

s = 3 m 

u = 0 

v = ? 

we know , 

F = ma 

5 x 10⁴ = 3 x 10⁷ x a 

5/3 x 10⁻³ = a 

1.6 x 10⁻³ = a 

we know 

v² = 2as + u²

v² = 2 x 1.6 x 10⁻³ x 3 + 0²

v² = 9.96 x 10⁻³ 

v = √ 9.96 x 10⁻³ 

v = 10⁻²(9.96 x 10)

v = 10⁻² x 9.97

v = 0.0997 m/s

Answer 2

Work done = force* distance = change in kinetic energy.

Initial K E = 0. Final KE = m v^2 /2.

v= final speed
50,000 * 3 = 1/2 * 3×10^7 * v^2
v^2 = 1/100
v = 0.1 m/s or 10 cm/sec
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Alternately find acceleration as force by mass and use equation v^2 = u^2 + 2 a s.