The expression for the acceleration of the cart for the experiment you did in lab is a = (m1g -...
a = (m1g - f)/(m1 + m2) where m1 + m2 is kept constant throughout.
(a) Consider the situation when friction force is zero. What is the expression for the acceleration of the cart now? (Use the following as necessary: m1, m2, f, andg.)
a =
If you now halve the hanging mass m1, (so m = 0.5m1), how will the value of the denominator for the expression for the acceleration change?
The denominator will halve in value. The denominator value will be m1/(2+m2). Halving m1 will increase m2 by that amount and the total mass will have a constant valueas required by the experiment.
What is the expression for the acceleration now? (Use the following as necessary: m1, m2, f, and g.)
a =
Based on your answers to the previous parts, what conclusion can you draw about the acceleration of the cart?
The cart's acceleration will halve as the numerator in the expression for the acceleration will halve when m1 is halved. The cart's acceleration will remain the sameas the total mass is constant during the experiment. The acceleration will increase by a certain value but not by a factor of ½.
(b) Consider the situation when friction force is non-zero. What is the expression for the acceleration of the cart now? (Use the following as necessary: m1, m2, f,and g.)
a =
If you now halve the hanging mass m1, (so m = 0.5m1) how will the value of the denominator for the expression for the acceleration change?
The denominator will halve in value. The denominator value will be m1/(2+m2). Halving m1 will increase m2 by that amount and the total mass will have a constant valueas required by the experiment.
What is the expression for the acceleration now? (Use the following as necessary: m1, m2, f, and g.)
a =
Based on your answers to the previous parts, what conclusion can you draw about the acceleration of the cart?
The cart's acceleration will halve as the numerator in the expression for the acceleration will halve when m1 is halved. The cart's acceleration will remain the sameas the total mass is constant during the experiment. The acceleration will increase by a certain value but not by a factor of ½.
Homework Answers
The expression for the acceleration of the cart for the experiment you did in lab is
a =
(m1g ? f)
(m1 + m2)
where
m1 + m2
is kept constant throughout.
Consider the situation when friction force is zero. What is the expression for the acceleration of the cart now? (Use the following as necessary: m1, m2, f, and g.)
a=(m_1 g)/(m_1+m_2 )
As the friction coefficient f will be zero.
If you now triple the hanging mass m1, (so m = 3m1), how will the value of the denominator for the expression for the acceleration change?
The denominator will triple in value.
The denominator value will be 3m1 + m2.
Tripling m1 will decrease m2 by that amount and the total mass will have a constant value as required by the experiment.
Answer: Tripling m1 will decrease m2 by that amount and the total mass will have a constant value as required by the experiment. As, experiment already mentioned that m_1+m_2 will be kept constant throughout.
What is the expression for the acceleration now? (Use the following as necessary: m1, m2, f, and g.)
a=(3m_1 g)/(?3m?_1+m_2 )
Based on your answers to the previous parts, what conclusion can you draw about the acceleration of the cart?
The cart's acceleration will triple as the numerator in the expression for the acceleration will triple when m1 is tripled.
The cart's acceleration will remain the same as the total mass is constant during the experiment.
The acceleration will increase by a certain value but not by a factor of 3.
Answer: The cart's acceleration will triple as the numerator in the expression for the acceleration will triple when m1 is tripled. And the denominator will adjust itself to remain same despite of tripling the mass m_1
(b) Consider the situation when friction force is non-zero. What is the expression for the acceleration of the cart now? (Use the following as necessary: m1, m2, f, and g.)
a=(m_1 g-f)/(m_1+m_2 )
If you now triple the hanging mass m1, (so m = 3m1) how will the value of the denominator for the expression for the acceleration change?
The denominator will triple in value
. The denominator value will be 3m1 + m2.
Tripling m1 will decrease m2 by that amount and the total mass will have a constant value as required by the experiment.
Answer: Tripling m1 will decrease m2 by that amount and the total mass will have a constant value as required by the experiment. As, experiment already mentioned that m_1+m_2 will be kept constant throughout.
What is the expression for the acceleration now? (Use the following as necessary: m1, m2, f, and g.)
a=(3m_1 g-f)/(?3m?_1+m_2 )
Based on your answers to the previous parts, what conclusion can you draw about the acceleration of the cart?
The cart's acceleration will triple as the numerator in the expression for the acceleration will triple when m1 is tripled
. The cart's acceleration will remain the same as the total mass is constant during the experiment.
The acceleration will increase by a certain value but not by a factor of 3.
Answer: The acceleration will increase by a certain value but not by a factor of 3.
a =
(m1g ? f)
(m1 + m2)
where
m1 + m2
is kept constant throughout.
Consider the situation when friction force is zero. What is the expression for the acceleration of the cart now? (Use the following as necessary: m1, m2,f, and g.)
a=(m_1 g)/(m_1+m_2 )
As the friction coefficient f will be zero.
If you now triple the hanging mass m1, (so m = 3m1), how will the value of the denominator for the expression for the acceleration change?
The denominator will triple in value.
The denominator value will be 3m1 + m2.
Tripling m1 will decrease m2 by that amount and the total mass will have a constant value as required by the experiment.
What is the expression for the acceleration now? (Use the following as necessary: m1, m2, f, and g.)
a=(3m_1 g-f)/(?3m?_1+m_2 )
Based on your answers to the previous parts, what conclusion can you draw about the acceleration of the cart?
The cart's acceleration will triple as the numerator in the expression for the acceleration will triple when m1 is tripled
. The cart's acceleration will remain the same as the total mass is constant during the experiment.
The acceleration will increase by a certain value but not by a factor of 3.
What is the expression for the acceleration now? (Use the following as necessary: m1, m2, f, and g.)
a=(3m_1 g)/(?3m?_1+m_2 )
Based on your answers to the previous parts, what conclusion can you draw about the acceleration of the cart?
The cart's acceleration will triple as the numerator in the expression for the acceleration will triple when m1 is tripled.
The cart's acceleration will remain the same as the total mass is constant during the experiment.
The acceleration will increase by a certain value but not by a factor of 3.
(b) Consider the situation when friction force is non-zero. What is the expression for the acceleration of the cart now? (Use the following as necessary:m1, m2, f, and g.)
a=(m_1 g-f)/(m_1+m_2 )
If you now triple the hanging mass m1, (so m = 3m1) how will the value of the denominator for the expression for the acceleration change?
The denominator will triple in value
. The denominator value will be 3m1 + m2.
Tripling m1 will decrease m2 by that amount and the total mass will have a constant value as required by the experiment.
Add Answer to:
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