Question

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 5.0 km/h. When a second tugboat applies an additional constant force of magnitude F2 in the same direction, the speed increases by 19 km/h during a 10 s interval. How do the magnitudes of F1 and F2 compare? (Neglect the effects of water resistance and air resistance.)

Answer 1

Answer

given,

time  = 10 s

ship's speed = 5 Km/h

F = m a

a is the acceleration and m is mass.

In the first case

F₁=m x a₁

where a₁ =  difference in velocity / time

F₁ is constant acceleration is also a constant.

Δv₁ = 5 x 0.278

Δv₁ = 1.39 m/s

$a_1=\dfrac{1.39}{10}$

a₁ = 0.139 m/s²

F₂ =m x a₂

F₃ = F₂ + F₁

Δv₃ = 19 x 0.278

Δv₃ = 5.282 m/s

a₃=Δv₂ / t

$a_3=\dfrac{5.282}{10}$

a₃ = 0.5282 m²/s

m a₃=m a₁ + m a₂

a₃ = a₂ + a₁

0.5282 = a₂ + 0.139

a₂=0.3892 m²/s

F₂ = m x 0.3892...........(1)

F₁ = m x 0.139...............(2)

F₂/F₁

ratio = $\dfrac{0.3892}{0.139}$

ratio = 2.8