chapter 5 forces in two dimensions practice problems
lesson 2 friction
example problem 3 - balanced friction forces
1. owen exerts a 36 - n horizontal force as she pulls a 52 - n sled across a cement sidewalk at constant speed. what is the coefficient of kinetic friction between the sidewalk and the metal sled runners? ignore air resistance.
2. mr. ames is dragging a box full of books from his office to his car. the box and books together have a combined weight of 134 n. if the coefficient of static friction between the pavement and the box is 0.55, how hard must mr. ames push horizontally on the box in order to start it moving?
3. thomas sits on a small rug on a polished wooden floor. the coefficient of kinetic friction between the rug and the slippery wooden floor is only 0.12. if thomas weighs 650 n, what horizontal force is needed to pull the rug and thomas across the floor at a constant speed?
4. you want to move a 41 - kg bookcase to a different place in the living room. if you push with a force of 65 n and the bookcase accelerates at 0.12 m/s², what is the coefficient of kinetic friction between the bookcase and the carpet?
5. consider the force pushing the box in example problem 4. how long would it take for the velocity of the box to double to 2.0 m/s?
6. ke min is driving at 23 m/s. he sees a tree branch lying across the road. he slams on the brakes when the branch is 60.0 m in front of him. if the coefficient of kinetic friction between the cars locked tires and the road is 0.41, will the car stop before hitting the branch? the car has a mass of 1200 kg.

Question

Accepted Answer

1.
# Explanation:
## Step1: Analyze forces at constant - speed
Since the sled is moving at a constant speed, the net force in the horizontal direction is zero. The horizontal force $F = 36N$ is equal to the kinetic - friction force $F_f$. The weight of the sled $W = 52N$, and the normal force $N$ on the sled is equal to its weight ($N = W$) because there is no acceleration in the vertical direction. The formula for kinetic - friction force is $F_f=\mu_kN$.
$F = F_f=\mu_kN$ and $N = W$, so $\mu_k=\frac{F}{N}=\frac{F}{W}$.
## Step2: Calculate the coefficient of kinetic friction
Substitute $F = 36N$ and $W = 52N$ into the formula: $\mu_k=\frac{36}{52}\approx0.69$.
# Answer:
$\mu_k\approx0.69$

2.
# Explanation:
## Step1: Determine the normal force
The combined weight of the box and books is $W = 134N$. The normal force $N$ on the box is equal to its weight ($N = W$) since there is no acceleration in the vertical direction.
## Step2: Calculate the static - friction force
The formula for static - friction force is $F_{s,max}=\mu_sN$. Substitute $\mu_s = 0.55$ and $N = 134N$ into the formula: $F_{s,max}=0.55	imes134 = 73.7N$. To start the box moving, the applied force must be at least equal to the maximum static - friction force.
# Answer:
$73.7N$

3.
# Explanation:
## Step1: Determine the normal force
The weight of Thomas is $W = 650N$. The normal force $N$ on the rug (which supports Thomas) is equal to his weight ($N = W$) since there is no acceleration in the vertical direction.
## Step2: Calculate the kinetic - friction force
The formula for kinetic - friction force is $F_f=\mu_kN$. Substitute $\mu_k = 0.12$ and $N = 650N$ into the formula: $F_f=0.12	imes650 = 78N$. Since the rug and Thomas are moving at a constant speed, the net force in the horizontal direction is zero, and the applied horizontal force $F$ is equal to the kinetic - friction force $F_f$.
# Answer:
$78N$

4.
# Explanation:
## Step1: Use Newton's second law in the horizontal direction
Newton's second law is $F_{net}=ma$. The net force in the horizontal direction is $F - F_f=ma$, where $F = 65N$, $m = 41kg$, and $a = 0.12m/s^2$. First, find the kinetic - friction force $F_f$: $F_f=F - ma$.
## Step2: Calculate the normal force
The normal force $N$ on the book - case is equal to its weight $N = mg$, where $g = 9.8m/s^2$ and $m = 41kg$, so $N=41	imes9.8 = 401.8N$.
## Step3: Calculate the coefficient of kinetic friction
The formula for kinetic - friction force is $F_f=\mu_kN$. Then $\mu_k=\frac{F_f}{N}$. Substitute $F_f=F - ma$ and $N = mg$ into the formula: $F_f=65-41	imes0.12=65 - 4.92 = 60.08N$. $\mu_k=\frac{60.08}{401.8}\approx0.15$.
# Answer:
$\mu_k\approx0.15$

5.
# Explanation:
## Step1: Use the kinematic equation
The kinematic equation $v = v_0+at$ is used. We know that $v_0$ (initial velocity) is not given, but assume it starts from rest $v_0 = 0m/s$, $v$ (final velocity) is $2.0m/s$, and $a = 0.12m/s^2$ (from the previous problem).
## Step2: Solve for time
Rearrange the kinematic equation $t=\frac{v - v_0}{a}$. Substitute $v = 2.0m/s$, $v_0 = 0m/s$, and $a = 0.12m/s^2$ into the formula: $t=\frac{2.0 - 0}{0.12}\approx16.67s$.
# Answer:
$t\approx16.67s$

6.
# Explanation:
## Step1: Calculate the normal force
The mass of the car is $m = 1200kg$. The normal force $N$ on the car is equal to its weight $N = mg$, where $g = 9.8m/s^2$, so $N = 1200	imes9.8=11760N$.
## Step2: Calculate the kinetic - friction force
The formula for kinetic - friction force is $F_f=\mu_kN$. Substitute $\mu_k = 0.41$ and $N = 11760N$ into the formula: $F_f=0.41	imes11760 = 4821.6N$.
## Step3: Use Newton's second law to find the deceleration
In the horizontal direction, $F_{net}=ma$. The net force is the kinetic - friction force $F_f$ (opposing the motion), so $-F_f=ma$. Solve for $a$: $a=-\frac{F_f}{m}=-\frac{4821.6}{1200}=-4.02m/s^2$.
## Step4: Use the kinematic equation to find the stopping distance
The kinematic equation $v^2 - v_0^2 = 2ax$ is used. The final velocity $v = 0m/s$, the initial velocity $v_0 = 23m/s$, and $a=-4.02m/s^2$. Rearrange the equation to solve for $x$: $x=\frac{v^2 - v_0^2}{2a}=\frac{0 - 23^2}{2	imes(-4.02)}=\frac{- 529}{-8.04}\approx65.8m$.
## Step5: Compare the stopping distance with the distance to the branch
Since $x\approx65.8m>60.0m$, the car will not stop before hitting the branch.
# Answer:
No

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze forces at constant - speed

Step2: Calculate the coefficient of kinetic friction

Step1: Determine the normal force

Step2: Calculate the static - friction force

Step1: Determine the normal force

Step2: Calculate the kinetic - friction force

Answer: