Question 1

The coefficient of friction between a racecar's wheels and the track is 1.0.The car starts from rest and accelerates at a constant rate for 700 m.Find the maximum speed at the end of the race.

Accepted Answer

To find the maximum speed, we need to use the formula: 
v^2 = 2ax 
where v is the final velocity, a is the acceleration and x is the distance covered. 
We know that the acceleration is constant and equal to the coefficient of friction times the acceleration due to gravity, which is 1*9.8 = 9.8 m/s^2.
So, plugging in the values, we get:
v^2 = 2*9.8*700 
v^2 = 13720 
v = √13720 
v = 117 m/s
Therefore, the correct choice is C.

Question 2

A 18-kg mass is placed on a 28° incline and friction keeps it from sliding.The coefficient of static friction in this case is 0.591,and the coefficient of sliding friction is 0.511.What is the frictional force in this situation?

Accepted Answer

The frictional force is equal to the component of the gravitational force parallel to the incline, which is m * g * sin(θ). For an 18-kg mass on a 28° incline, this is 18 * 9.8 * sin(28°) = 82.81 N.

Question 3

A 300-N sled is pulled up a 27° slope at a constant speed by a force of 250 N.What is the coefficient of kinetic friction between sled and slope?

Accepted Answer

The answer of A 300-N sled is pulled up a...

Question 4

Hector drives a pickup truck horizontally at 10.0 m/s.He is transporting a crate of delicate lead crystal.If the coefficient of static friction between the crate and the truck bed is 0.700,what is the minimum stopping distance for the truck so the crate will not slide?

Accepted Answer

The minimum stopping distance can be calculated using the formula $ d = \frac{v^2}{2\mu g} $, where $ v = 10.0 \, \text{m/s} $, $ \mu = 0.700 $, and $ g = 9.8 \, \text{m/s}^2 $. Plugging in these values gives $ d = \frac{(10.0)^2}{2 \times 0.700 \times 9.8} = 7.3 \, \text{m} $.

Question 5

Dana uses a rope to pull a box that weighs 300 N across a level surface with constant velocity.The rope makes an angle of 25° above the horizontal and the tension in the rope is 100 N.What is the coefficient of friction?

Accepted Answer

The horizontal component of the tension is 100 N * cos(25°). The frictional force equals this horizontal component since the box moves with constant velocity. The normal force is the weight minus the vertical component of the tension (100 N * sin(25°)). The coefficient of friction is the frictional force divided by the normal force, which calculates to approximately 0.35.

Question 6

A 12-kg mass is placed on a 29° incline and friction keeps it from sliding.The coefficient of static friction in this case is 0.633,and the coefficient of sliding friction is 0.524.The mass is given a shove causing it to slide down the incline.What is the frictional force while the mass is sliding?

Accepted Answer

The frictional force while the mass is sliding is calculated using the coefficient of sliding friction. The normal force is the component of the gravitational force perpendicular to the incline, which is $mg \cos(\theta)$. Thus, the frictional force is $ \mu_k \times mg \cos(\theta) = 0.524 \times 12 \times 9.8 \times \cos(29^\circ) \approx 53.90 \, \text{N} $.

Question 7

Jamal pulls a 250-N sled up a 25.0° slope at constant speed by a force of 200 N.Near the top of the hill he releases the sled.With what acceleration does the sled go down the hill?

Accepted Answer

Since the sled is moving at constant speed up the slope, it must be experiencing a force that balances the force of gravity acting on it down the slope. Therefore, we can set up the following equation:

force pulling sled up slope = force of gravity down slope

200 N = mgsin(25°)

Solving for the mass of the sled, we get:

m = 200 N / (9.8 m/s^2) * sin(25°) = 44.22 kg

When Jamal releases the sled, the only force acting on it is the force of gravity down the slope. Using Newton's second law (force = mass x acceleration), we can solve for the acceleration of the sled down the slope:

force of gravity = ma

mg*sin(25°) = ma

a = g*sin(25°) = 0.44 m/s^2

Therefore, the sled accelerates down the slope at a rate of 0.44 m/s^2.

Question 8

Two objects,A and B,are placed on an inclined plane that can be rotated to different angles of elevation.A starts to slide at twice the angle of elevation that B starts sliding.The respective coefficients for static friction for A and B are μ_A and μ_B.Choose the last answer that is correct.

Accepted Answer

Let the angle at which object B starts sliding be θ. 
Then, the angle at which object A starts sliding is 2θ. 
The force of static friction is given by F_s = μ_sN, where N is the normal force. 
Since the objects are on an inclined plane, the normal force is given by N = m*g*cos(θ), where m is the mass of the object and g is the acceleration due to gravity. 
The force of gravity acting on the objects is given by F_g = m*g*sin(θ).

For object B: 
Frictional force, F_fB = μS1U1B1BN = μS1U1B1BS1U1B0 * m*g*cos(θ) 
Gravitational force, F_gB = m*g*sin(θ)

For object A: 
Frictional force, F_fA = μS1U1B1AN = μS1U1B1AS1U1B0 * m*g*cos(2θ) 
Gravitational force, F_gA = m*g*sin(2θ)

To determine which object will start sliding first, we need to compare the frictional force to the gravitational force acting on the objects:

For object B: 
F_fB = μS1U1B1BS1U1B0 * m*g*cos(θ) 
F_gB = m*g*sin(θ) 
When F_fB = F_gB, object B will start sliding.

For object A: 
F_fA = μS1U1B1AS1U1B0 * m*g*cos(2θ) 
F_gA = m*g*sin(2θ) 
When F_fA = F_gA, object A will start sliding.

Setting F_fB = F_gB: 
μS1U1B1BS1U1B0 * m*g*cos(θ) = m*g*sin(θ) 
μS1U1B1BS1U1B0 = tan(θ)

Setting F_fA = F_gA: 
μS1U1B1AS1U1B0 * m*g*cos(2θ) = m*g*sin(2θ) 
μS1U1B1AS1U1B0 = tan(2θ)/2

Since tangent is an increasing function in the interval [0, π/2), we have tan(2θ)/2 > tan(θ), which means that μS1U1B1AS1U1B0 > μS1U1B1BS1U1B0. Answer: B

Question 9

A trapeze artist,with swing,weighs 800 N;he is being held to one side by his partner so that the swing ropes make an angle of 45.0° with the vertical.In such a condition of static equilibrium,what is the tension in the rope?

Accepted Answer

The tension in the rope can be found using the vertical component of the tension, which must equal the weight of the trapeze artist. T * cos(45°) = 800 N, solving for T gives approximately 1,131 N.

Question 10

A hockey puck moving at 9.0 m/s coasts to a halt in 85 m on a smooth ice surface.What is the coefficient of friction between the ice and the puck?

Accepted Answer

We can use the formula: 
d = (v^2/2μg)
where d is the distance traveled, v is the initial velocity, μ is the coefficient of friction and g is the acceleration due to gravity (9.8 m/s^2)
Rearranging the formula to solve for μ, we get: 
μ = (v^2)/(2gd)
Substituting the given values, we get: 
μ = (9.0^2)/(2 x 9.8 x 85) ≈ 0.049
Therefore, the coefficient of friction between the ice and the puck is 0.049. So, the best choice is B.

Question 11

A trapeze artist,with swing,weighs 600 N;he is momentarily held to one side by his partner so that the swing ropes make an angle of 45° with the vertical.In such a condition of static equilibrium,what is the horizontal force being applied by the partner?

Accepted Answer

The answer of A trapeze artist,with swing,weighs 600 N;he is...

Question 12

A man pulls a sled at a constant velocity across a horizontal snow surface.If a force of 85 N is being applied to the sled rope at an angle of 50° to the ground,what is the force of friction between sled and snow?

Accepted Answer

The answer of A man pulls a sled at a...

Question 13

A worker pulls a 350-N packing crate at constant velocity across a rough floor by exerting a force F = 200.0 N at an angle of 30.0° above the horizontal.What is the coefficient of kinetic friction of the floor?

Accepted Answer

The coefficient of kinetic friction ($\mu_k$) can be found using the equation $F_{friction} = \mu_k F_{normal}$. Since the crate moves at constant velocity, the net force is zero, meaning the horizontal component of the applied force equals the frictional force, and the vertical component of the applied force plus the normal force equals the weight of the crate. The horizontal component of the applied force is $F \cos(\theta) = 200 \cos(30^\circ) = 173.2 \, \text{N}$, and the vertical component is $F \sin(\theta) = 200 \sin(30^\circ) = 100 \, \text{N}$. The normal force is the weight of the crate minus the vertical component of the applied force, $350 \, \text{N} - 100 \, \text{N} = 250 \, \text{N}$. Thus, $\mu_k = \frac{F_{friction}}{F_{normal}} = \frac{173.2}{250} = 0.693$.

Question 14

As a car moves forward on a level road at constant velocity,the net force acting on the tires is:

Accepted Answer

According to Newton's first law of motion, an object moving at a constant velocity must have a net force of zero acting on it. Therefore, the net force acting on the tires of a car moving at a constant velocity on a level road must be zero.

Question 15

As a car skids with its wheels locked trying to stop on a road covered with ice and snow,the force of friction between the icy road and the tires will usually be:

Accepted Answer

When the wheels are locked, the skid results in kinetic friction, which is less than the static friction. Therefore, the force of friction between the icy road and the tires will be less than the normal force of the road times the coefficient of static friction (μs).

Question 16

There are five books in a stack,each with a weight of 8N.The coefficient of friction between all the books is 0.4 as is the coefficient between the table and the bottom book.What horizontal push must I just exceed on the next to bottom book to start sliding the top four books off the bottom one?

Accepted Answer

The force of friction that needs to be overcome to slide the top four books off the bottom one is calculated as: Friction force = coefficient of friction * normal force. The normal force here is the weight of the top four books, which is 4 books * 8N = 32N. Therefore, the friction force = 0.4 * 32N = 12.8N.

Question 17

A 479-N crate rests on an ramp;the maximum angle just before it slips is 20° with the horizontal.What is the coefficient of static friction between crate and ramp surfaces?

Accepted Answer

The answer of A 479-N crate rests on an ramp;the...

Question 18

A 17-kg mass is placed on a 28° incline and friction keeps it from sliding.The coefficient of static friction in this case is 0.565 and the coefficient of sliding friction is 0.515.The mass is given a shove causing it to slide down the incline.Taking down the incline as positive,what is the acceleration of the mass while it is sliding?

Accepted Answer

The answer of A 17-kg mass is placed on a...

Question 19

As a car goes up a hill,there is a force of friction between the road and the tires rolling on the road.The maximum force of friction is equal to:

Accepted Answer

The maximum force of friction is equal to the product of the normal force and the coefficient of static friction. In this case, as the car is going up a hill, it is not slipping or sliding, instead it is pushing against the road with a force equal to its weight, which is counteracted by the normal force of the road. Therefore, the maximum force of friction is equal to the product of the normal force and the coefficient of static friction.

Question 20

An Olympic skier moving at 15.0 m/s down a 25.0° slope encounters a region of wet snow,of coefficient of friction μ_k = 0.740.How far down the slope does she go before stopping?

Accepted Answer

The skier's initial kinetic energy is converted into work done against friction. Using the equation $d = \frac{v^2}{2g(\sin\theta + \mu_k \cos\theta)}$, where $v = 15.0 \, \text{m/s}$, $g = 9.8 \, \text{m/s}^2$, $\theta = 25.0^\circ$, and $\mu_k = 0.740$, the distance $d$ is calculated to be approximately 46 m.

Quiz 4: The Laws of Motion