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Test BGK
HW#10
Question: A pinball machine's plunger has a spring constant of 22
N/m and is compressed by 0.04 m to start a 0.006 kg pinball.
1. What is the elastic potential energy before the ball is released?
2. What is the kinetic energy of the pinball the instant it leaves the
spring?
3. What is the speed of the pinball the instant it leaves the spring?
4. If the pinball is moving at 1.3 m/s as it is deflected horizontally
across the top of the pinball machine, how much higher above the
ground is this part of its path when compared to its starting
position?
Squation
Drawing
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Signature
Instructions.
Answers
Answer:
1. The elastic potential energy is 0.0176 Joules
2. The kinetic energy of the pinball the instant it leaves the spring is 0.0176 Joules
3. The speed of the pinball the instant it leaves the spring is approximately 2.42212 m/s
4. The height of the part where the pinball is located on the machine above the ground is approximately 0.213 meters
Explanation:
The spring constant of the pinball machine's plunger, k = 22 N/m
The amount by which the pinball machine's plunger is compressed, x = 0.04 m
The mass of the pinball ball, m = 0.006 kg
1. The elastic potential energy, P.E. = 1/2·k·x²
By substitution, we get;
P.E. = 1/2 × 22 N/m × (0.04 m)² = 0.0176 J
The elastic potential energy, P.E. = 0.0176 J
2. At the instant the pinball leaves the spring, the plunger and therefore the force of the plunger no longer acts on the pinball
Since there are no external forces acting on the pinball to increase the speed of the pinball after it leaves the spring, the velocity reached is its maximum velocity, and therefore, the kinetic energy, K.E. is the maximum kinetic energy which by the conservation of energy, is equal to the initial potential energy
Therefore;
K.E. = P.E. = 0.0176 J
The kinetic energy of the pinball the instant it leaves the spring, K.E.= 0.0176 J
3. The kinetic energy, K.E., is given by the following formula;
K.E. = 1/2·m·v²
Where;
v = The speed or velocity of the object having kinetic energy K.E.
Therefore, from K.E. = 0.0176 J, and by plugging in the values of the variables, we have;
K.E. = 0.0176 J = 1/2 × 0.006 kg × v²
v² = 0.0176 J/(1/2 × 0.006 kg) = 88/15 m²/s²
v = √(88/15 m²/s²) ≈ (2·√330)/15 m/s ≈ 2.42212 m/s
The speed of the pinball the instant it leaves the spring, v ≈ 2.42212 m/s
4. The height of the pinball is given by the following kinematic equation of motion;
[tex]v_h[/tex]² = u² - 2·g·h
Where;
[tex]v_h[/tex] = The velocity of the pinball at the given height = 1.3 m/s
u = v ≈ 2.42212 m/s (The initial velocity of the pinball as it the spring)
g = The acceleration due to gravity ≈ 9.8 m/s²
h = The height of the pinball above the ground
We get;
[tex]v_h[/tex]² = 1.3² = 2.42212² - 2 × 9.8 × h
∴ h = (2.42212² - 1.3²)/(2 × 9.8) ≈ 0.213
The height of the part where the pinball is located on the machine above the ground, h ≈ 0.213 m
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Explanation:
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Hope this helps have a awesome night/day❤️✨
Explanation:
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(Your question never states if both objects are unique or similar polar charges, so I just assumed they were both neutral objects existing within an electric field.)
So a better question would then be, what is gravity’s relationship with an electric field?
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PLEASE HELLP ILL GIVE BrAINLY AND HEART
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Answer:
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i hope this helps
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West
East
<
디
50 Newtons
40 Newtons
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B
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that's my Answer
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Answer:
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yw .
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