Acceleration is rate of change of A-PositionB-TimeC-VelocityD-Speed
2 answers:
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Answer:
a) 0.28 m or 28 cm is the minimum height above ground the fish reaches.
b) at the height of 0.484 m height , the pufferfish will eventually come to rest.
c) There exists two types of energy remain at the equilibrium point in the system. These are :
Gravitational potential energy = 23.72J
Spring potential energy = 0.384 J
Explanation:
Given that :
Mass of the pufferfish m =5kg
initial height of the fish h =5.5m
length of the spring l =0.5m
Spring constant K =3000N/m
a)
Assuming no energy loss to friction, what is the minimum height above the ground that the pufferfish reaches?
Lets assume that the minimum height the fish reaches is = x meters
Now by using the conservation of energy; we realize that :
Initial total energy = final total energy
Gravitational potential energy =
Gravitational potential energy' + Spring potential energy (kinetic energy is zero in both cases)
Replacing our given values into the above equation; we have :
269.5 = 47.5 x + 1500(0.5 -x )²
269.5 = 47.5 x + 1500(0.25 - x²)
269.5 = 47.5 x + 375 - 1500 x²
269.5 - 375 = 47.5 x - 1500 x²
-105.5 = 47.5 x - 1500 x²
-105.5 + 1500 x² - 47.5 x = 0
1500 x² - 47.5 x - 105.5 = 0
By using quadratic equation and taking the positive value;
x = 0.28 m or 28 cm is the minimum height above ground the fish reaches.
b)
At the equilibrium position the weight of fish will be equal to the force applied by the spring thus
mg = kx
substituting our given values ; we have:
(5)(9.8) = 3000x
x = 61.22
x = 0.016m : so this is the compression in the spring
Now; to determine the height the pufferfish gets to before it eventually come to rest; we have
(0.5-0.016) m = 0.484m
therefore, at the height of 0.484 m height , the pufferfish will eventually come to rest.
c)
There exists two types of energy remain at the equilibrium point in the system. These are :
Gravitational potential energy = mgh' = (5)(9.8)(0.484)
= 23.72J
and spring potential energy
Answer:
- Our final answer is 15 kg .
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