A marble is placed at the bottom of a semi-spherical bowl, as shown in the figure. The marble is then displaced from the bottom
of the bowl to a position about halfway from the top of the bowl. The marble is then released from rest such that the marble always remains in contact with the bowl. Students observe that the marble rolls back and forth as it oscillates about the bottom of the bowl. Which of the following statements best explains why the marble undergoes oscillatory motion?
a)The sides of the bowl become steeper at positions farther from the bottom of the bowl.b) The net force exerted on the marble always has a component directed toward the bottom of the bowl. c) The normal force exerted on the marble decreases with increasing distance from the bottom of the bowl. D ) The gravitational force exerted on the marble is constant in magnitude and direction.
a)The sides of the bowl become steeper at positions farther from the bottom of the bowl.b) The net force exerted on the marble always has a component directed toward the bottom of the bowl. c) The normal force exerted on the marble decreases with increasing distance from the bottom of the bowl. D ) The gravitational force exerted on the marble is constant in magnitude and direction.
1 answer:
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Explanation:
- Newton's first law of motion:
"An object at rest (or in uniform motion) remains at rest (or in uniform motion) unless acted upon an unbalanced force
In this situation, we can apply Newton's first law to the keys of the keyboard that are not hit by the fingers of the man. In fact, as no force act on the keys, they remain at rest.
- Newton's second law of motion:
"The acceleration experienced by an object is proportional to the net force exerted on the object; mathematically:
where F is the net force, m is the mass of the object, and a its acceleration"
In this case, we can apply Newton's second law to the keys of the keyboard that are hit by the man: in fact, as they are hit, they experience a downward force, and therefore they experience a downward acceleration.
"Newton's third law of motion:
"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"
Here We can apply Newton's third law to the pair of objects finger-key: in fact, as the finger apply a force on the key (action force), then the key exerts a force back on the finger (reaction force), equal and opposite.
Answer:
Amplitude = 0.058m
Frequency = 6.25Hz
Explanation:
Given
Amplitude (A) = 8.26 x 10-2 m
Frequency (f) = 4.42Hz
Conversation of energy before split
½mv² = ½KA²
Make A the subject of formula
A =
Conversation of energy after split
½(m/2)V'² = ½(m/2)V² = ½KA'²
½(m/2)V² = ½KA'²
Make A the subject of formula
First divide both sides by ½
(m/2)V² = KA'²
Divide both sides by K
V² = A'²
= A'
Substitute for A in the above equation
A' = A/√2
A' = 8.26 x 10^-2/√2
A' = 0.05840702012600882
Amplitude after split = 0.058 (Approximated)
Frequency (f') = f√2
f' = 4.42√2
f' = 6.25082394568908011
Frequency after split = 6.25Hz (approximated)