22) the free body diagram shown below represents a crate being dragged towards the left over a rough surface. which of the vectors represent the normal force and the friction force acting on the crate?
table and diagram omitted
23) an unfortunate bug splatters against the windshield of a moving car. compared to the force of the car on the bug, the force of the bug on the car is
a) smaller b) the same c) larger d)need more information to decide
24) another unfortunate bug splatters against the windshield of a moving car. compared to the deceleration of the car, the deceleration of the bug is
a) less b) the same c) greater d) need more information to decide
25) john is pulling a 40.0 kg cart with a horizontal force of 100 n. the force of friction of the cart is 20.0 n. the acceleration of the cart is
a) 2.00 m/s²
b) 2.50 m/s²
c) 20.0 m/s²
d) 200 m/s²
26) the force of friction experienced by a 76.5 kg cross-country skier sliding on unwaxed hickory skis along dry snow at a constant speed of 4.00 m/s is -60.0 n. what is the coefficient of friction between unwaxed hickory skis and dry snow?
diagram of skier on incline omitted

Question

22) the free body diagram shown below represents a crate being dragged towards the left over a rough surface. which of the vectors represent the normal force and the friction force acting on the crate?
table and diagram omitted
23) an unfortunate bug splatters against the windshield of a moving car. compared to the force of the car on the bug, the force of the bug on the car is
a) smaller  b) the same  c) larger  d)need more information to decide
24) another unfortunate bug splatters against the windshield of a moving car. compared to the deceleration of the car, the deceleration of the bug is
a) less  b) the same  c) greater  d) need more information to decide
25) john is pulling a 40.0 kg cart with a horizontal force of 100 n. the force of friction of the cart is 20.0 n. the acceleration of the cart is
a) 2.00 m/s²
b) 2.50 m/s²
c) 20.0 m/s²
d) 200 m/s²
26) the force of friction experienced by a 76.5 kg cross-country skier sliding on unwaxed hickory skis along dry snow at a constant speed of 4.00 m/s is -60.0 n. what is the coefficient of friction between unwaxed hickory skis and dry snow?
diagram of skier on incline omitted

Accepted Answer

### Question 22
# Explanation:
## Step1: Recall Normal and Friction Forces
Normal force acts perpendicular to the surface (upward for a crate on a horizontal surface), friction force opposes motion (since crate moves left, friction is right? Wait, no—wait, the crate is dragged left, so friction opposes, so friction is right? Wait, no, the free - body diagram: $ \vec{F}_1 $ is left, $ \vec{F}_3 $ is right, $ \vec{F}_2 $ is up, $ \vec{F}_4 $ is down. Normal force is perpendicular to surface (up, so $ \vec{F}_2 $), friction force opposes motion (crate moves left, so friction is right? Wait, no, if the crate is being dragged left, the friction force should act to the right? Wait, no, wait: when an object is moving left, the kinetic friction force is opposite to the direction of motion, so right. But in the diagram, $ \vec{F}_1 $ is left, $ \vec{F}_3 $ is right, $ \vec{F}_2 $ is up (normal force, perpendicular to surface), $ \vec{F}_4 $ is down (gravity). Wait, the normal force is the force from the surface on the crate, so it's upward ($ \vec{F}_2 $), and friction force: since the crate is moving left, friction is right? Wait, no, maybe I got the direction wrong. Wait, the crate is being dragged left, so the applied force is left, and friction opposes, so friction is right. But in the options, let's check:

Normal force: perpendicular to surface (so vertical, upward: $ \vec{F}_2 $). Friction force: parallel to surface, opposite to motion (motion is left, so friction is right? Wait, no, the diagram has $ \vec{F}_1 $ left, $ \vec{F}_3 $ right, $ \vec{F}_2 $ up, $ \vec{F}_4 $ down. Wait, maybe the crate is moving left, so friction is right? No, wait, if the crate is moving left, the friction force is to the right (opposing motion). But in the table:

Option C: Normal force $ \vec{F}_2 $, Friction force $ \vec{F}_4 $? No, wait $ \vec{F}_4 $ is down. Wait, maybe I messed up. Wait, normal force is upward ($ \vec{F}_2 $), friction force: since the surface is rough and the crate is moving left, friction is to the right? No, $ \vec{F}_1 $ is left, $ \vec{F}_3 $ is right. Wait, maybe the applied force is left, friction is right ($ \vec{F}_3 $)? No, the normal force is $ \vec{F}_2 $ (upward). Let's re - analyze:

Normal force: acts perpendicular to the contact surface, so for a crate on a horizontal surface, normal force is upward (so $ \vec{F}_2 $). Friction force: acts parallel to the surface, opposing the direction of motion. The crate is moving left, so friction force is to the right? Wait, no, if the crate is being dragged left, the friction force is to the right (opposing the motion). But in the diagram, $ \vec{F}_1 $ is left, $ \vec{F}_3 $ is right, $ \vec{F}_2 $ is up, $ \vec{F}_4 $ is down. Wait, maybe the answer is C: Normal force $ \vec{F}_2 $, Friction force $ \vec{F}_1 $? No, that can't be. Wait, maybe I have the direction of friction wrong. If the crate is moving left, the friction force is to the right? No, no—friction opposes relative motion. If the crate is moving left, the surface exerts a friction force to the right on the crate. Wait, but in the options:

Option C: Normal force $ \vec{F}_2 $, Friction force $ \vec{F}_4 $? No, $ \vec{F}_4 $ is down. Wait, maybe the diagram is misinterpreted. Let's look at the options again. The normal force should be vertical (upward), so $ \vec{F}_2 $. The friction force should be horizontal, opposing motion. The crate is moving left, so friction is right? But $ \vec{F}_3 $ is right. Wait, no, maybe the crate is being dragged left, so the friction is left? No, that would be in the same direction. No, friction opposes motion. So if the crate is moving left, friction is right. So normal force is $ \vec{F}_2 $, friction force is $ \vec{F}_3 $? But option B is normal force $ \vec{F}_2 $, friction force $ \vec{F}_3 $? Wait, the table:

Option A: Normal $ \vec{F}_1 $, Friction $ \vec{F}_2 $

Option B: Normal $ \vec{F}_2 $, Friction $ \vec{F}_3 $

Option C: Normal $ \vec{F}_2 $, Friction $ \vec{F}_4 $

Option D: Normal $ \vec{F}_4 $, Friction $ \vec{F}_1 $

Ah, I see. Normal force is upward ($ \vec{F}_2 $), friction force: since the crate is moving left, friction is to the right ($ \vec{F}_3 $)? Wait, no, $ \vec{F}_1 $ is left, $ \vec{F}_3 $ is right. Wait, maybe the applied force is left ($ \vec{F}_1 $), and friction is right ($ \vec{F}_3 $), normal is up ($ \vec{F}_2 $), gravity is down ($ \vec{F}_4 $). So normal force is $ \vec{F}_2 $, friction force is $ \vec{F}_3 $? But option B is normal $ \vec{F}_2 $, friction $ \vec{F}_3 $? Wait, the question says "dragged towards the left", so the direction of motion is left, so friction is right (opposing). So normal force is $ \vec{F}_2 $ (upward), friction force is $ \vec{F}_3 $ (rightward). So option B? Wait, no, maybe I made a mistake. Wait, the normal force is perpendicular to the surface, so upward ($ \vec{F}_2 $), friction force is parallel to the surface, opposing motion (motion is left, so friction is right, $ \vec{F}_3 $). So the answer is B? Wait, no, let's check again.

Wait, the free - body diagram: $ \vec{F}_1 $ is left (maybe applied force), $ \vec{F}_3 $ is right (maybe friction? No, friction should oppose motion. If the crate is moving left, friction is right. So if the crate is being dragged left, the applied force is left, friction is right. So normal force is up ($ \vec{F}_2 $), friction is right ($ \vec{F}_3 $). So option B: Normal $ \vec{F}_2 $, Friction $ \vec{F}_3 $.

## Step2: Confirm the Forces
Normal force: perpendicular to the surface (vertical, upward: $ \vec{F}_2 $). Friction force: parallel to the surface, opposite to the direction of motion (motion is left, so friction is right: $ \vec{F}_3 $). So the correct option is B.

# Answer: B. $ \vec{F}_2 $ (Normal Force), $ \vec{F}_3 $ (Friction Force)

### Question 23
# Brief Explanations:
According to Newton's third law of motion, when two objects interact, the force exerted by object A on object B is equal in magnitude and opposite in direction to the force exerted by object B on object A. In this case, the car and the bug are interacting objects. The force of the car on the bug and the force of the bug on the car are action - reaction pairs. So they have the same magnitude.

# Answer: B) the same

### Question 24
# Brief Explanations:
We know from Newton's second law $ F = ma $, or $ a=\frac{F}{m} $. The force exerted on the bug and the car are equal in magnitude (Newton's third law). Let $ F $ be the magnitude of the force. The mass of the bug ($ m_{bug} $) is much smaller than the mass of the car ($ m_{car} $). Since $ a=\frac{F}{m} $, and $ F $ is the same for both, a smaller mass will result in a larger acceleration. So the deceleration of the bug (which is the acceleration of the bug in the opposite direction of its motion) is greater than the deceleration of the car.

# Answer: C) greater

### Question 25
# Explanation:
## Step1: Identify the Net Force
The net force $ F_{net} $ acting on the cart is the horizontal applied force minus the friction force. The applied force $ F = 100\ N $ (horizontal), and the friction force $ f = 20.0\ N $ (opposing the motion, so in the opposite direction of the applied force). So $ F_{net}=F - f $.
$ F_{net}=100\ N- 20.0\ N = 80\ N $

## Step2: Use Newton's Second Law to Find Acceleration
Newton's second law is $ F_{net}=ma $, where $ m = 40.0\ kg $ is the mass of the cart, and $ a $ is the acceleration. We can solve for $ a $ by rearranging the formula: $ a=\frac{F_{net}}{m} $
Substitute $ F_{net} = 80\ N $ and $ m = 40.0\ kg $ into the formula: $ a=\frac{80\ N}{40.0\ kg}=2.00\ m/s^{2} $

# Answer: A) $ 2.00\ m/s^{2} $

### Question 26
# Explanation:
## Step1: Analyze the Forces on the Skier
The skier is moving at a constant speed, so the net force along the vertical direction (perpendicular to the slope) is zero. The normal force $ F_n $ and the component of the gravitational force perpendicular to the slope balance each other. The gravitational force $ F_g=mg $, where $ m = 76.5\ kg $ and $ g = 9.8\ m/s^{2} $. The angle of the slope is $ 30^{\circ} $, so the component of the gravitational force perpendicular to the slope is $ F_{g\perp}=mg\cos\theta $, and since the skier is in equilibrium in the perpendicular direction, $ F_n = mg\cos\theta $

## Step2: Calculate the Normal Force
First, calculate $ mg $: $ mg=76.5\ kg\times9.8\ m/s^{2}=749.7\ N $
Then, $ \cos(30^{\circ})=\frac{\sqrt{3}}{2}\approx0.866 $
So $ F_n=mg\cos\theta = 749.7\ N\times0.866\approx649.2\ N $

## Step3: Use the Friction Force Formula to Find the Coefficient of Friction
The formula for kinetic friction is $ f = \mu_kF_n $, where $ f=- 60.0\ N $ (the negative sign indicates direction, we can use the magnitude $ |f| = 60.0\ N $), and $ \mu_k $ is the coefficient of kinetic friction. We can solve for $ \mu_k $: $ \mu_k=\frac{|f|}{F_n} $
Substitute $ |f| = 60.0\ N $ and $ F_n\approx649.2\ N $ into the formula: $ \mu_k=\frac{60.0\ N}{649.2\ N}\approx0.0924 $ (approximate value, depending on the value of $ g $ used. If we use $ g = 9.81\ m/s^{2} $):

$ mg = 76.5\ kg\times9.81\ m/s^{2}=750.465\ N $

$ F_n=750.465\ N\times\cos(30^{\circ})\approx750.465\times0.8660\approx649.9\ N $

$ \mu_k=\frac{60.0\ N}{649.9\ N}\approx0.0923 $

# Answer: Approximately $ 0.092 $ (or more precisely, using $ g = 9.8\ m/s^{2} $, $ \mu_k\approx0.0924 $; using $ g = 9.81\ m/s^{2} $, $ \mu_k\approx0.0923 $)

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QUESTION IMAGE

Question

Explanation:

Question 22

Step1: Recall Normal and Friction Forces

Answer:

Question 23