# Homework For Lab 3 Force And Motion Answer Key

Homework for Lab 3: Force and Motion Answer Key

## Homework for Lab 3: Force and Motion Answer Key

If you are looking for the answer key to the homework for lab 3 on force and motion, you have come to the right place. In this article, we will go over the questions and solutions for this lab assignment, which covers topics such as Newton's laws of motion, vectors, and friction. We will also provide some tips and resources to help you master these concepts and ace your physics course.

## Question 1

You are given 10 identical springs of force (i.e., a means of producing repeatable forces of a variety of sizes). Describe how you would use an uncalibrated force probe and the springs in question 1 to develop a quantitative scale of force. How could you measure forces that do not correspond to exact numbers of stretched springs?

Download File: __https://miimms.com/2w39rL__

### Solution

To develop a quantitative scale of force, you could use the following steps:

Attach one end of a spring to a fixed point and the other end to the force probe.

Pull the spring until it is stretched by a certain amount (e.g., 10 cm) and record the reading on the force probe. This will give you the force exerted by one spring stretched by that amount.

Repeat step 2 with different numbers of springs (e.g., 2, 3, 4, etc.) attached in series to the force probe. Record the readings for each case. This will give you the forces exerted by different numbers of springs stretched by the same amount.

Plot a graph of force vs. number of springs. The slope of this graph will give you the force per spring stretched by that amount.

Use the equation F = kx, where F is the force, k is the spring constant (the slope of the graph), and x is the amount of stretch, to calculate the force for any given stretch.

To measure forces that do not correspond to exact numbers of stretched springs, you could use interpolation or extrapolation on the graph. For example, if you want to measure the force exerted by 2.5 springs stretched by 10 cm, you could find the point on the graph that corresponds to 2.5 on the x-axis and read off the y-value. Alternatively, you could use the equation F = kx with x = 2.5 * 10 cm.

## Question 2

What is meant by a proportional relationship? Is this the same as a linear relationship? Explain. How would you determine if the relationship between widgets and doodads is a proportional one? Sketch on the axes to the right of the table what the graph would look like if widgets are proportional to doodads.

WidgetsDoodads

0.0150.5

10.0305.0

20.0442.7

30.0601.3

40.0742.9

### Solution

A proportional relationship is one in which two quantities vary directly with each other, meaning that as one quantity increases or decreases, so does the other quantity by the same factor. A proportional relationship can be represented by an equation of the form y = kx, where y and x are the two quantities, and k is a constant called the proportionality constant.

A linear relationship is one in which two quantities vary linearly with each other, meaning that as one quantity increases or decreases by a fixed amount, so does the other quantity by a fixed amount. A linear relationship can be represented by an equation of the form y = mx + b, where y and x are the two quantities, m is a constant called the slope, and b is a constant called the y-intercept.

A proportional relationship is a special case of a linear relationship, where b = 0. This means that a proportional relationship passes through the origin (0,0) on a graph, while a linear relationship may or may not pass through the origin depending on the value of b.

To determine if the relationship between widgets and doodads is a proportional one, we could check if the ratio of doodads to widgets is constant for all values in the table. If the ratio is constant, then the relationship is proportional and can be written as doodads = k * widgets, where k is the proportionality constant. If the ratio is not constant, then the relationship is not proportional and may be linear or nonlinear.

From the table, we can calculate the ratio of doodads to widgets for each row as follows:

WidgetsDoodadsRatio

0.0150.5Undefined

10.0305.030.5

20.0442.722.135

30.0601.320.043

40.0742.918.572

We can see that the ratio is not constant, and in fact decreases as widgets increase. Therefore, the relationship between widgets and doodads is not proportional, and we cannot use a straight line that passes through the origin to represent it on a graph.

A possible sketch of the graph of doodads vs. widgets is shown below:

## Question 3

A cart can move along a horizontal line (the + position axis). It moves with the velocity shown below.

What is happening to the cart's position at each of these times: t = 1 s, t = 3 s, t = 4 s?

What is happening to the cart's velocity at each of these times: t = 1 s, t = 3 s, t = 4 s?

Roughly sketch the position-time graph for the cart on the axes below.

### Solution

The cart's position at each time can be determined by looking at the area under the velocity-time graph up to that time. The area represents the displacement (change in position) of the cart from its initial position. The sign of the area indicates the direction of the displacement (positive for rightward, negative for leftward). The magnitude of the area indicates the size of the displacement.

At t = 1 s, the area under the graph is a positive rectangle with base 1 s and height 2 m/s. The area is 2 m^2, which means that the cart has moved 2 m to the right from its initial position.

At t = 3 s, the area under the graph is a positive rectangle with base 3 s and height 2 m/s, plus a negative triangle with base 2 s and height -2 m/s. The area is 6 m^2 - 2 m^2 = 4 m^2, which means that the cart has moved 4 m to the right from its initial position.

At t = 4 s, the area under the graph is zero, which means that the cart has returned to its initial position.

The cart's velocity at each time can be determined by looking at the value of the velocity-time graph at that time. The sign of the value indicates the direction of the velocity (positive for rightward, negative for leftward). The magnitude of the value indicates the speed of the cart.

At t = 1 s, the value of the graph is 2 m/s , which means that the cart is moving to the right with a speed of 2 m/s.

At t = 3 s, the value of the graph is 0 m/s, which means that the cart is momentarily at rest.

At t = 4 s, the value of the graph is -2 m/s, which means that the cart is moving to the left with a speed of 2 m/s.

To sketch the position-time graph for the cart, we can use the following steps:

Choose a suitable scale for the x-axis (time) and the y-axis (position).

Mark the initial position of the cart at t = 0 s on the y-axis. This will be the starting point of the graph.

Use the information from part a to plot the position of the cart at t = 1 s, t = 3 s, and t = 4 s. These will be some points on the graph.

Use the information from part b to draw straight lines connecting the points. The slope of each line segment will represent the velocity of the cart in that time interval. A positive slope means a positive velocity (rightward), a negative slope means a negative velocity (leftward), and a zero slope means a zero velocity (at rest).

A possible sketch of the position-time graph for the cart is shown below:

## Question 4

A block of mass 2 kg is placed on a rough horizontal surface. A horizontal force of 12 N is applied to the block. The block accelerates at 3 m/s^2. What is the coefficient of kinetic friction between the block and the surface?

### Solution

To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass times its acceleration. In this case, the net force acting on the block is the applied force minus the frictional force. The frictional force is proportional to the normal force, which is equal to the weight of the block in this case. The proportionality constant is called the coefficient of kinetic friction, denoted by mu_k. Therefore, we can write:

F_net = F - F_f = ma

F_f = mu_k * F_N = mu_k * mg

Substituting these equations, we get:

12 - mu_k * 2 * 9.8 = 2 * 3

Solving for mu_k, we g