Question

name: wyaneca
date: oct 9 2015 period: 1
relationship between force, mass and acceleration analysis

1. look at your graphed line for experiment 1. explain why it does or does not indicate that there is a relationship between force and acceleration.
2. what is the mathematical relationship between force, mass and acceleration?
3. compare the two lines, \experiment 1\ and \experiment 2\ on your graph. identify and explain:

a. any similarities
b. any differences

1. fill in the table below using your equation for force, mass, and acceleration to find the missing values.

experiment 3 force (n)	mass of block (kg)	acceleration of block (m/s²)
20	2	10
10	5	2
50	10	5
100	4	25
1,000	40	25

1. in the first activity, vehicle 2 has a greater acceleration than vehicle 1, but has a less forceful engine. how can this be? explain in terms of your equation.
2. one newton of force is the same as 1kg m/s². explain how this unit of measurement is appropriate for your equation.

Explanation:

Step1: Answer question 1

If the graphed line for Experiment 1 is a straight - line passing through the origin (or has a non - zero slope), it indicates a direct proportional relationship between force and acceleration. According to Newton's second law \(F = ma\) (where \(F\) is force, \(m\) is mass and \(a\) is acceleration), when mass is constant, force and acceleration are directly proportional. If the line is horizontal or vertical or a random curve not showing a clear trend, it may not indicate a relationship.

Step2: Answer question 2

The mathematical relationship is \(F=ma\), where \(F\) is the net force acting on an object, \(m\) is the mass of the object, and \(a\) is the acceleration of the object.

Step3: Answer question 3a

Similarities: If both lines are straight and pass through the origin, it means in both experiments, when mass is considered constant, force and acceleration follow a direct proportional relationship as per \(F = ma\).

Step4: Answer question 3b

Differences: The slopes of the two lines may be different. The slope of the \(F - a\) graph is equal to the mass (\(m=\frac{F}{a}\)). A steeper slope indicates a larger mass, while a less steep slope indicates a smaller mass.

Step5: Answer question 4

Using the formula \(F = ma\), we can find the missing values. For example, if \(F = 25N\) and \(m = 5kg\), then \(a=\frac{F}{m}=\frac{25}{5}=5m/s^{2}\).

Step6: Answer question 5

From \(F = ma\), we can express \(a=\frac{F}{m}\). Vehicle 2 has a greater acceleration with a less forceful engine. This means that the mass of Vehicle 2 must be less than the mass of Vehicle 1. A smaller mass will result in a larger acceleration for a given force.

Step7: Answer question 6

From \(F = ma\), the unit of mass \(m\) is kilogram (\(kg\)) and the unit of acceleration \(a\) is meters per second squared (\(m/s^{2}\)). When we multiply mass and acceleration (\(kg\times m/s^{2}\)), we get the unit of force, which is Newton (\(N\)). So, \(1N = 1kg\cdot m/s^{2}\) is consistent with the formula \(F = ma\).

Answer:

1. Depends on the shape of the graphed line. If it's a straight - line through the origin (or with non - zero slope), there is a relationship; otherwise, there may not be.
2. \(F = ma\)
3. a. If both are straight lines through the origin, force and acceleration are directly proportional in both. b. Different slopes mean different masses.
4. Calculated using \(F = ma\) (values filled in the table as per calculations).
5. Vehicle 2 has a smaller mass, as \(a=\frac{F}{m}\).
6. Since \(F = ma\), and mass is in \(kg\) and acceleration in \(m/s^{2}\), the unit of force is \(kg\cdot m/s^{2}\), which is a Newton.