How Does Subtracting and Adding Masses Affect Acceleration in a Force Equation?

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In summary, the given force produces accelerations of 11.9 m/s^2 for an object with mass m1 and 2.8 m/s^2 for an object with mass m2. To find the acceleration for an object with mass m2-m1, the equation a = F/(m2-m1) can be used, where F is equal to 11.9*m1 and 2.8*m2. Similarly, the acceleration for an object with mass m2+m1 can be found using the same equation, but with F equal to 11.9*m1 and 2.8*m2. In the provided solution, an error was made when using m2-m1 instead of
  • #1
mcdowellmg
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Homework Statement



A certain force gives an object of mass m1 an acceleration of 11.9 m/s^2 and an object of mass m2 an acceleration of 2.8 m/s^2. What acceleration would the force give to an object of mass

m2 - m1

and

m2 + m1?

Homework Equations



F=ma
Force = mass*acceleration

The Attempt at a Solution



I know that a = F/(m2-m1).
11.9*m1 = F and 2.8*m2 = F, so I know that 11.9*m1 = 2.8*m2, because F is the same throughout.
Therefore, m1= F/11.9 and m2= F/2.8. Now m1- m2= F/11.9- F/2.8= -0.273109F.
This would lead me to believe that -0.273109F = m1-m2, so if a=F/(m2-m1), then a should be F/-0.273109F, which is -3.661.

However, that is not coming up as correct. I haven't attempted the m2+m1 mass, because I am trying to figure out the first one, and when I do, it should be easy to compute.EDIT- Well, never mind. Apparently, it was 3.66, without the negative. Where did I go wrong with the negative?
 
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  • #2
mcdowellmg said:

Homework Statement



A certain force gives an object of mass m1 an acceleration of 11.9 m/s^2 and an object of mass m2 an acceleration of 2.8 m/s^2. What acceleration would the force give to an object of mass

m2 - m1

and

m2 + m1?

Homework Equations



F=ma
Force = mass*acceleration

The Attempt at a Solution



I know that a = F/(m2-m1).
11.9*m1 = F and 2.8*m2 = F, so I know that 11.9*m1 = 2.8*m2, because F is the same throughout.
Therefore, m1= F/11.9 and m2= F/2.8. Now m1- m2= F/11.9- F/2.8= -0.273109F.
This would lead me to believe that -0.273109F = m1-m2, so if a=F/(m2-m1), then a should be F/-0.273109F, which is -3.661.
Start with:

[tex]F/a_1 = m_1 \text{ and } F/a_2 = m_2[/tex]

From that work out m2-m1 and m1+m2 in terms of F, a1 and a2. Then find the accelerations for m2-m1 and m2+m1

AM
 
  • #3
In the problem they have asked the acceleration of m2 - m1 which is equal to 3,661 m/s^2
 
  • #4
Andrew Mason said:
Start with:

[tex]F/a_1 = m_1 \text{ and } F/a_2 = m_2[/tex]

From that work out m2-m1 and m1+m2 in terms of F, a1 and a2. Then find the accelerations for m2-m1 and m2+m1

AM

the -0.273109F u got is m1-m2,and u used it as m2-m1 to find the acceleration.its just a careless error noting more
 
  • #5

Your approach to solving the problem is correct, but there is a mistake in your final calculation. When you substitute the values for m1 and m2 into the equation a = F/(m2-m1), you should get a = F/(-0.273109F). The negative sign should be in the denominator, not in front of the entire fraction. This gives you a = -3.66 m/s^2, which is the correct answer.

For the m2+m1 mass, you can follow the same approach and substitute the values for m1 and m2 into the equation a = F/(m2+m1). This will give you a = F/(14.7m).
 

Related to How Does Subtracting and Adding Masses Affect Acceleration in a Force Equation?

What is the formula for calculating force (F) using mass (m) and acceleration (a)?

The formula for calculating force (F) is F = m x a. This means that force is equal to the mass of an object multiplied by its acceleration.

Why is the Basic F=ma Problem considered tricky?

The Basic F=ma Problem is considered tricky because it involves understanding and applying the concept of force, mass, and acceleration in a real-world situation. It requires critical thinking and problem-solving skills to determine the correct values for each variable and solve the equation.

How do I approach solving a Basic F=ma Problem?

To solve a Basic F=ma Problem, you should first identify the given information, including the values for mass, acceleration, and force. Then, plug these values into the formula F = m x a and solve for the missing variable. It is important to pay attention to units and use the correct units in the equation.

What are some common mistakes to avoid when solving a Basic F=ma Problem?

Common mistakes when solving a Basic F=ma Problem include using incorrect units, plugging in the wrong values, and forgetting to account for the direction of force or acceleration. It is important to double-check your work and make sure all values and units are correct before arriving at a final answer.

Can the Basic F=ma Problem be applied to real-world situations?

Yes, the Basic F=ma Problem can be applied to real-world situations. It is commonly used in physics and engineering to calculate the force needed to move an object, the acceleration of an object in motion, and other related calculations. Understanding this concept is crucial in many scientific and technological fields.

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