To solve this problem we will apply the concepts given by the kinematic equations of motion. For this purpose it will be necessary with the given data to obtain the deceleration. With this it will be possible again to apply one of the kinematic equations of motion that does not depend on time, but on distance, to find how far the block would slide with the quadruplicate velocity
Our values are given as,
Using the kinematic equation of motion we have
Now if the initial velocity is quadrupled we have that,
Replacing the values
Therefore the block would have slipped around 30.44 if its initial velocity quadrupled.
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Answer:</h2>
105146 Pa
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Explanation:</h2>
1) We will make a Free-Body Diagram representing all the upward and downward pressures exerted on the piston.
- Pressure exerted by the compressed spring (Pspring)
- Pressure due to weight of the piston (Pw)
- Atmospheric pressure (Patm)
- Initial pressure inside the cylinder. (P1)
2) We will formulate an equation balancing all upward and downward pressures.
P1= Patm + Pw + Pspring
3) We will calculate each of the pressures separately.
P = F/A
F= ks
k= 38×1000 =38000 N m
s= 2.5 /1000 = (2.5x10^-3) m
F = 38000×(2.5x10^-3) = 95 N
A = 30/10000 = (30x10^-4) m2
P = 95 / (30x10^-4)
Pspring ≅ 3167 Pa
P = F/A
F = W = mg
W = 2×9.81 = 19.62 N
A = 30/10000 = (30x10^-4) m2
P = 19.62 / (30x10^-4)
Pw = 654 Pa
P = 1atm = 101325 Pa
Patm = 101325 Pa
4) We will add all the downward pressures to reach the final answer (initial pressure inside the cylinder).
P1= Patm + Pw + Pspring
P1= 101325+654+3167
P1= 105146 Pa
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Answer:
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