ws#2 - friction
directions: draw a fbd and show your ( f_{ ext{net}} ) equation for all. put all answers to 2 sig digs.
1. it takes a 50 n horizontal force to pull a 20 kg object along the ground at a constant velocity. what is the coefficient of friction?
2. if the coefficient of friction is 0.30, how much horizontal force is needed to pull a mass of 15 kg across a level board at a uniform velocity?
3. it takes 775 n of force to keep a ford truck moving at a constant velocity with a coefficient of friction of 0.31. what is its mass?
4. the coefficient of friction between rubber and ice is 0.040. how much force is needed to keep a 175 g puck sliding?
5. a 5.0 kg box will accelerate at 2.0 m/s² when we push it with 29.6 n of force. what is the coefficient of friction?
6. a cart with a mass of 2.0 kg is pulled across a level desk by a force horizontally of 4.0 n. if the coefficient of kinetic friction is 0.12, what is the acceleration of the cart?
7. a 4.8 kg object is accelerating across a level surface with a coefficient of friction of 0.40. the force of friction is 15 n and the applied force is 25 n. what is the acceleration?

answers
1. ( mu = 0.26 )
2. ( f_a = 44 , ext{n} )
3. ( m = 260 , ext{kg} )
4. ( f_a = 0.069 , ext{n} )
5. ( mu = 0.40 )
6. ( a = 0.82 , ext{m/s}^2
ightarrow ( ext{right}) )
7. ( a = 2.1 , ext{m/s}^2
ightarrow ( ext{right}) )

Question

ws#2 - friction
directions: draw a fbd and show your ( f_{	ext{net}} ) equation for all. put all answers to 2 sig digs.
1. it takes a 50 n horizontal force to pull a 20 kg object along the ground at a constant velocity. what is the coefficient of friction?
2. if the coefficient of friction is 0.30, how much horizontal force is needed to pull a mass of 15 kg across a level board at a uniform velocity?
3. it takes 775 n of force to keep a ford truck moving at a constant velocity with a coefficient of friction of 0.31. what is its mass?
4. the coefficient of friction between rubber and ice is 0.040. how much force is needed to keep a 175 g puck sliding?
5. a 5.0 kg box will accelerate at 2.0 m/s² when we push it with 29.6 n of force. what is the coefficient of friction?
6. a cart with a mass of 2.0 kg is pulled across a level desk by a force horizontally of 4.0 n. if the coefficient of kinetic friction is 0.12, what is the acceleration of the cart?
7. a 4.8 kg object is accelerating across a level surface with a coefficient of friction of 0.40. the force of friction is 15 n and the applied force is 25 n. what is the acceleration?

answers
1. ( mu = 0.26 )
2. ( f_a = 44 , 	ext{n} )
3. ( m = 260 , 	ext{kg} )
4. ( f_a = 0.069 , 	ext{n} )
5. ( mu = 0.40 )
6. ( a = 0.82 , 	ext{m/s}^2
ightarrow (	ext{right}) )
7. ( a = 2.1 , 	ext{m/s}^2
ightarrow (	ext{right}) )

Accepted Answer

### Problem 1:
# Explanation:
## Step1: Identify forces (constant velocity, so $ F_{net} = 0 $)
Horizontal force $ F = 50 \, \text{N} $, friction $ f = \mu N $. Normal force $ N = mg $ (since level surface), $ m = 20 \, \text{kg} $, $ g = 9.8 \, \text{m/s}^2 $. So $ F = f = \mu mg $.
## Step2: Solve for $ \mu $
$ \mu = \frac{F}{mg} = \frac{50}{20 \times 9.8} $
$ \mu = \frac{50}{196} \approx 0.255 \approx 0.26 $ (2 sig digs)
# Answer: $ \boldsymbol{\mu = 0.26} $

### Problem 2:
# Explanation:
## Step1: Friction force formula ($ f = \mu N $), $ N = mg $
$ m = 15 \, \text{kg} $, $ \mu = 0.30 $, $ g = 9.8 \, \text{m/s}^2 $. Constant velocity, so applied force $ F = f $.
## Step2: Calculate $ F $
$ F = \mu mg = 0.30 \times 15 \times 9.8 $
$ F = 0.30 \times 147 = 44.1 \approx 44 \, \text{N} $ (2 sig digs)
# Answer: $ \boldsymbol{F = 44 \, \text{N}} $

### Problem 3:
# Explanation:
## Step1: $ F = f = \mu N $, $ N = mg $, so $ F = \mu mg $
Solve for $ m $: $ m = \frac{F}{\mu g} $
## Step2: Substitute values ($ F = 775 \, \text{N} $, $ \mu = 0.31 $, $ g = 9.8 \, \text{m/s}^2 $)
$ m = \frac{775}{0.31 \times 9.8} = \frac{775}{3.038} \approx 255 \approx 260 \, \text{kg} $ (2 sig digs)
# Answer: $ \boldsymbol{m = 260 \, \text{kg}} $

### Problem 4:
# Explanation:
## Step1: Convert mass to kg: $ m = 175 \, \text{g} = 0.175 \, \text{kg} $
Normal force $ N = mg $, friction $ f = \mu N $, constant velocity so $ F = f $.
## Step2: Calculate $ F $
$ F = \mu mg = 0.040 \times 0.175 \times 9.8 $
$ F = 0.040 \times 1.715 = 0.0686 \approx 0.069 \, \text{N} $ (2 sig digs)
# Answer: $ \boldsymbol{F = 0.069 \, \text{N}} $

### Problem 5:
# Explanation:
## Step1: $ F_{net} = F_{applied} - f = ma $
Friction $ f = \mu N = \mu mg $. So $ F_{applied} - \mu mg = ma $
## Step2: Solve for $ \mu $
$ \mu = \frac{F_{applied} - ma}{mg} $
Substitute $ F_{applied} = 29.6 \, \text{N} $, $ m = 5.0 \, \text{kg} $, $ a = 2.0 \, \text{m/s}^2 $, $ g = 9.8 \, \text{m/s}^2 $
$ \mu = \frac{29.6 - (5.0 \times 2.0)}{5.0 \times 9.8} = \frac{29.6 - 10}{49} = \frac{19.6}{49} = 0.40 $
# Answer: $ \boldsymbol{\mu = 0.40} $

### Problem 6:
# Explanation:
## Step1: $ F_{net} = F_{applied} - f $, $ f = \mu N = \mu mg $
$ m = 2.0 \, \text{kg} $, $ F_{applied} = 4.0 \, \text{N} $, $ \mu = 0.12 $, $ g = 9.8 \, \text{m/s}^2 $
## Step2: Calculate $ F_{net} $ then $ a $
$ f = 0.12 \times 2.0 \times 9.8 = 2.352 \, \text{N} $
$ F_{net} = 4.0 - 2.352 = 1.648 \, \text{N} $
$ a = \frac{F_{net}}{m} = \frac{1.648}{2.0} \approx 0.82 \, \text{m/s}^2 $ (2 sig digs)
# Answer: $ \boldsymbol{a = 0.82 \, \text{m/s}^2} $

### Problem 7:
# Explanation:
## Step1: $ F_{net} = F_{applied} - f $
$ F_{applied} = 25 \, \text{N} $, $ f = 15 \, \text{N} $, so $ F_{net} = 25 - 15 = 10 \, \text{N} $
## Step2: Calculate acceleration ($ a = \frac{F_{net}}{m} $), $ m = 4.8 \, \text{kg} $
$ a = \frac{10}{4.8} \approx 2.083 \approx 2.1 \, \text{m/s}^2 $ (2 sig digs)
# Answer: $ \boldsymbol{a = 2.1 \, \text{m/s}^2} $

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QUESTION IMAGE

Question

Explanation:

Problem 1:

Step1: Identify forces (constant velocity, so \( F_{net} = 0 \))

Step2: Solve for \( \mu \)

Step1: Friction force formula (\( f = \mu N \)), \( N = mg \)

Step2: Calculate \( F \)

Step1: \( F = f = \mu N \), \( N = mg \), so \( F = \mu mg \)

Step2: Substitute values (\( F = 775 \, \text{N} \), \( \mu = 0.31 \), \( g = 9.8 \, \text{m/s}^2 \))

Answer:

Problem 2: