Standing and holding a barbell that has a mass of 100kg at a height of 2m involves being done on the barbell to maintain
1 answer:
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Explanation:
Since, the rod is present in vertical position and the spring is unrestrained.
So, initial potential energy stored in the spring is = 0
And, initial potential gravitational potential energy of the rod is .
It is given that,
mass of the bar = 0.795 kg
g = 9.8
L = length of the rod = 0.2 m
Initial total energy T =
Now, when the rod is in horizontal position then final total energy will be as follows.
T =
where, I = moment of inertia of the rod about the end =
Also,
where, = speed of the tip of the rod
x = spring extension
The initial unstrained length is = 0.1 m
Therefore, final length will be calculated as follows.
x' = m
Then, x =
x = m - 0.1 m
= 0.1236 m
k = 25 N/m
So, according to the law of conservation of energy
Putting the given values into the above formula as follows.
v = 2.079 m/s
Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.
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