7. a 7.26kg bowling ball (16 pounds) is at rest at the end of a bowling lane. use this information to answer the following questions.
a. what is the net force acting on the ball?
b. you push the ball with a force of 22.8n which induces a -2.3n frictional force. what is the net force while you push the ball?
c. what is the acceleration caused by your push on the ball?
d. following your push the ball rolls down the lane at 4.2m/s. what is the net force on the ball as it rolls down the lane at the constant speed?
e. challenge: how long did you push the ball in this situation?
exercises
8. complete the following table. be sure to include units in your answer.
| net force (n) | mass (kg) | acceleration |
|---------------|-----------|--------------|
| 5.0 | 2.5 | |
| 2.32 | 12 | |
| 18.2 | 1.967 | |
| 0.87 | 21.2 | |
| 180 | 1.793 | |
9. complete the following table. be sure to include units.
| net force (n) | mass (kg) | acceleration (m/s²) |
|---------------|-----------|---------------------|
| | 4.0 | 4.0 |
| 25.0 | | 4.998 |
| 53 | 3 | |
| 172.44 | | 4.665 |
| | 1225 | 3.43 |

Question

7. a 7.26kg bowling ball (16 pounds) is at rest at the end of a bowling lane. use this information to answer the following questions.
   a. what is the net force acting on the ball?
   b. you push the ball with a force of 22.8n which induces a -2.3n frictional force. what is the net force while you push the ball?
   c. what is the acceleration caused by your push on the ball?
   d. following your push the ball rolls down the lane at 4.2m/s. what is the net force on the ball as it rolls down the lane at the constant speed?
   e. challenge: how long did you push the ball in this situation?
exercises
8. complete the following table. be sure to include units in your answer.
| net force (n) | mass (kg) | acceleration |
|---------------|-----------|--------------|
| 5.0           | 2.5       |              |
| 2.32          | 12        |              |
| 18.2          | 1.967     |              |
| 0.87          | 21.2      |              |
| 180           | 1.793     |              |
9. complete the following table. be sure to include units.
| net force (n) | mass (kg) | acceleration (m/s²) |
|---------------|-----------|---------------------|
|               | 4.0       | 4.0                 |
| 25.0          |           | 4.998               |
| 53            | 3         |                     |
| 172.44        |           | 4.665               |
|               | 1225      | 3.43                |

Accepted Answer

### Question 7a
# Explanation:
## Step1: Recall Newton's first law
An object at rest has a net force of 0 because $ F_{net} = ma $, and if $ v = 0 $ (at rest) and not accelerating ($ a = 0 $), then $ F_{net}=m\times0 = 0 $.
# Answer:
The net force acting on the ball is $ 0\,\text{N} $.

### Question 7b
# Explanation:
## Step1: Sum the forces
Net force is the sum of applied force and frictional force. $ F_{net}=F_{applied}+F_{friction} $
## Step2: Substitute values
$ F_{applied} = 22.8\,\text{N} $, $ F_{friction}=- 2.3\,\text{N} $, so $ F_{net}=22.8+( - 2.3)=20.5\,\text{N} $
# Answer:
The net force while pushing the ball is $ 20.5\,\text{N} $.

### Question 7c
# Explanation:
## Step1: Recall Newton's second law
$ F = ma $, so $ a=\frac{F_{net}}{m} $
## Step2: Substitute values
$ F_{net} = 20.5\,\text{N} $, $ m = 7.26\,\text{kg} $, so $ a=\frac{20.5}{7.26}\approx2.82\,\text{m/s}^2 $
# Answer:
The acceleration is approximately $ 2.82\,\text{m/s}^2 $.

### Question 7d
# Explanation:
## Step1: Recall Newton's first law for constant speed
If an object moves at constant speed, its acceleration $ a = 0 $. From $ F_{net}=ma $, if $ a = 0 $, then $ F_{net}=m\times0=0 $
# Answer:
The net force on the ball is $ 0\,\text{N} $.

### Question 7e
# Explanation:
## Step1: Recall kinematic equation $ v = u+at $
Initial velocity $ u = 0\,\text{m/s} $, final velocity $ v = 4.2\,\text{m/s} $, acceleration $ a\approx2.82\,\text{m/s}^2 $ (from part c)
## Step2: Solve for time $ t $
$ t=\frac{v - u}{a}=\frac{4.2 - 0}{2.82}\approx1.49\,\text{s} $
# Answer:
The time for which the ball was pushed is approximately $ 1.49\,\text{s} $.

### Question 8
We use $ a=\frac{F}{m} $ for each row:

1. $ F = 5.0\,\text{N} $, $ m = 2.5\,\text{kg} $: $ a=\frac{5.0}{2.5}=2.0\,\text{m/s}^2 $
2. $ F = 2.32\,\text{N} $, $ m = 12\,\text{kg} $: $ a=\frac{2.32}{12}\approx0.193\,\text{m/s}^2 $
3. $ F = 18.2\,\text{N} $, $ m = 1.967\,\text{kg} $: $ a=\frac{18.2}{1.967}\approx9.25\,\text{m/s}^2 $
4. $ F = 0.87\,\text{N} $, $ m = 21.2\,\text{kg} $: $ a=\frac{0.87}{21.2}\approx0.041\,\text{m/s}^2 $
5. $ F = 180\,\text{N} $, $ m = 1.793\,\text{kg} $: $ a=\frac{180}{1.793}\approx100.4\,\text{m/s}^2 $

| Net Force (N) | Mass (kg) | Acceleration ($ \text{m/s}^2 $) |
|---------------|-----------|-----------------------------------|
| 5.0           | 2.5       | 2.0                               |
| 2.32          | 12        | 0.193                             |
| 18.2          | 1.967     | 9.25                              |
| 0.87          | 21.2      | 0.041                             |
| 180           | 1.793     | 100.4                             |

### Question 9
We use $ F = ma $ (for finding force) and $ m=\frac{F}{a} $ (for finding mass) and $ a=\frac{F}{m} $ (for finding acceleration):

1. $ m = 4.0\,\text{kg} $, $ a = 4.0\,\text{m/s}^2 $: $ F=4.0\times4.0 = 16.0\,\text{N} $
2. $ F = 25.0\,\text{N} $, $ a = 4.998\,\text{m/s}^2 $: $ m=\frac{25.0}{4.998}\approx5.00\,\text{kg} $
3. $ F = 53\,\text{N} $, $ m = 3\,\text{kg} $: $ a=\frac{53}{3}\approx17.7\,\text{m/s}^2 $
4. $ F = 172.44\,\text{N} $, $ a = 4.665\,\text{m/s}^2 $: $ m=\frac{172.44}{4.665}\approx37.0\,\text{kg} $
5. $ m = 1225\,\text{kg} $, $ a = 3.43\,\text{m/s}^2 $: $ F=1225\times3.43 = 4201.75\,\text{N} $

| Net Force (N) | Mass (kg) | Acceleration ($ \text{m/s}^2 $) |
|---------------|-----------|-----------------------------------|
| 16.0          | 4.0       | 4.0                               |
| 25.0          | 5.00      | 4.998                             |
| 53            | 3         | 17.7                              |
| 172.44        | 37.0      | 4.665                             |
| 4201.75       | 1225      | 3.43                              |

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QUESTION IMAGE

Question

Explanation:

Question 7a

Step1: Recall Newton's first law

Step1: Sum the forces

Step2: Substitute values

Step1: Recall Newton's second law

Step2: Substitute values

Answer:

Question 7b