# Why is my momentum equation giving an incorrect force?

• k_squared
In summary, the conversation discusses two boxes, Box A and Box B, with weights of 100 pounds and 50 pounds respectively, and a coefficient of kinetic friction of 0.25. A 50 pound force is pushing Box A into Box B, and the question asks for their speed at 5 seconds and the force that Box B exerts on Box A. The correct answer for the force is found by subtracting the force of friction from the net force acting on Box B.
k_squared

## Homework Statement

There are two boxes, sitting right next to each other. Call the box on the left "box A" and the box on the right "Box B". A 50 pound force starts pushing A into B. Find their speed at 5 seconds, and how much force Box B puts on Box A. Box A weighs 100 pounds, Box B is 50 pounds and the coefficient of kinetic friction is 0.25.

## Homework Equations

(mv)_1 + (integral of force with repsect to time) = (mv)_2

## The Attempt at a Solution

First, I solved for their speed and acceleration using basic equations of motion (It is in fact 13.4 ft/s. This implies an acceleration of 2.68 ft/s.) Then, I solved for the force that Box B places on box A. My attempt goes like this:

F_(b on a) * 5 = (50/32.2)*13.4. This yields and answer of 4.1 lbs. However, by doing F_(b on a)-(50(0.25))=(50/32.2)(2.68) I get the correct answer, (because of Newton's third law.)

Why does my first equation not yield the correct answer? Thanks for any help!

k_squared said:
Why does my first equation not yield the correct answer?

Is your F_(b on a) the only force acting on block B? The change in momentum (or impulse) will be due to the net force acting on the block.

## 1. Why is my momentum equation not balancing?

There are a few possible reasons why your momentum equation may not be balancing. One common reason is that there may be a mistake in your calculations or assumptions. Double check all of your inputs and equations to make sure they are correct. Another possibility is that there may be external forces acting on the system that you have not accounted for in your equation. Finally, your system may not be in equilibrium, which can result in an unbalanced momentum equation.

## 2. Why is the force calculated by my momentum equation different from the actual force?

If the force calculated by your momentum equation is different from the actual force, it is likely due to errors in your measurements or assumptions. Make sure you are using accurate and precise data in your calculations. Additionally, your momentum equation may not be taking into account all of the forces acting on the system, such as friction or air resistance. Consider revising your equation to account for these additional forces.

## 3. How can I determine which term in my momentum equation is causing the incorrect force?

To determine which term in your momentum equation is causing the incorrect force, you can try isolating each term and calculating its individual contribution to the overall force. This can help identify any errors or discrepancies in your equation. You may also want to check your units to make sure they are consistent throughout the equation.

## 4. Can the momentum equation give incorrect forces in certain situations?

Yes, the momentum equation can give incorrect forces in certain situations. One example is when the system is not isolated and there are external forces acting on it. In these cases, the momentum equation will not accurately describe the forces at play. Additionally, if the system is not in equilibrium or if there are errors in the inputs or assumptions, the momentum equation may give incorrect forces.

## 5. How can I improve the accuracy of my momentum equation?

To improve the accuracy of your momentum equation, make sure you are using precise and accurate data in your calculations. Double check all of your inputs and assumptions to ensure they are correct. If possible, try to isolate and account for all of the forces acting on the system. You may also want to consider using more advanced equations or models that take into account additional factors such as friction and air resistance.

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