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
1 answer:
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
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
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 =
ratio = 2.8
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A)
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Parallel:
(1)
where
m is the mass
g = 9.8 m/s^2 the acceleration of gravity
is the coefficient of friction
R is the normal reaction
a is the acceleration
Perpendicular:
(2)
From (2) we find
And substituting into (1)
Solving for a,
B) 5.94 m/s
We can solve this part by using the suvat equation
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
Here we have
v = ?
u = 0 (it starts from rest)
s = 8.70 m
Solving for v,