Question 1

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 given by the product of the coefficient of static friction and the normal force of the road. This is because when the car is moving up the hill, it is not in a state of constant motion, but rather a state of equilibrium, hence static friction comes into play. The weight of the car does not directly affect the maximum force of friction, although it does affect the normal force of the road. Also, kinetic friction only applies once the car is already in motion, not when it is trying to overcome static friction to start moving.

Question 2

A block is launched up an incline plane.After going up the plane,it slides back down to its starting position.The coefficient of friction between the block and the plane is 0.3.The speed of the block when it reaches the starting position on the trip down:

Accepted Answer

The speed of the block when it reaches the starting position on the trip down will be less than the launching speed due to the presence of friction. As the block goes up the incline, some of its initial kinetic energy is converted into potential energy. When it comes back down, it will have less potential energy than it did when it was launched, which means it will have less kinetic energy and hence a lower speed. The frictional force acting on the block as it slides down the plane further reduces its speed. Therefore, option B is the best choice.

Question 3

A horizontal force of 850 N is needed to overcome the force of static friction between a level floor and a 350-kg crate.If g = 9.8 m/s²,what is the coefficient of static friction?

Accepted Answer

The equation for static friction is:

f_s = μ_s N

where f_s is the force of static friction, μ_s is the coefficient of static friction, and N is the normal force (equal in magnitude to the force of gravity).

In this problem, the force of static friction is given as 850 N and the mass of the crate is 350 kg.

The normal force can be found using:

N = m*g

N = 350 kg * 9.8 m/s^2

N = 3430 N

Substituting these values into the equation for static friction, we get:

850 N = μ_s * 3430 N

μ_s = 850 N / 3430 N

μ_s = 0.248

Therefore, the coefficient of static friction is 0.248, which corresponds to choice D.

Question 4

Hector drives a pickup truck horizontally at 9.00 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 0.740,what is the minimum stopping distance for the truck so the crate will not slide?

Accepted Answer

To prevent the crate from sliding, the maximum acceleration (a) must not exceed the acceleration due to friction, which can be calculated using the formula $a = \mu g$, where $\mu$ is the coefficient of static friction (0.740) and $g$ is the acceleration due to gravity (9.8 m/s$^2$). Thus, $a = 0.740 \times 9.8 = 7.252$ m/s$^2$. Using the equation $v^2 = u^2 + 2as$ (where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $s$ is the distance), and setting $v = 0$ for a complete stop, we solve for $s$: $0 = (9.00)^2 + 2(-7.252)s$. Solving for $s$ gives $s = 5.58$ m, which means the minimum stopping distance to prevent the crate from sliding is 5.58 meters.

Question 5

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 of elevation at which B starts sliding be θ. Then, A starts sliding at 2θ. The force of static friction is given by fs ≤ μsN, where N is the normal force acting on the object. The normal force is given by N = mgcosθ, where m is the mass of the object and g is the acceleration due to gravity. The force component parallel to the inclined plane is given by mg sinθ for object B and mg sin(2θ) for object A. For object B, the force of static friction is fs ≤ μsBmgcosθ. The condition for sliding to start is fs = mg sinθ, i.e. μsBmgcosθ = mg sinθ, which gives μsB = tanθ. For object A, the force of static friction is fs ≤ μsAmgcos2θ. The condition for sliding to start is fs = mg sin(2θ), i.e. μsAmgcos2θ = mg sin(2θ), which gives μsA = tanθ/2. Therefore, μsB > μsA, or μ_A > μ_B.

Question 6

A 400-N crate rests on a ramp;the maximum angle just before it slips is 15.0° with the horizontal.What is the coefficient of static friction between crate and ramp surfaces?

Accepted Answer

The coefficient of static friction ($\mu_s$) can be calculated using the formula $\mu_s = \tan(\theta)$, where $\theta$ is the angle of the incline. For $\theta = 15.0°$, $\mu_s = \tan(15.0°) \approx 0.27$.

Question 7

A sled is on a horizontal surface and a rope with tension T is pulling on it at an angle θ above the horizontal.The sled had mass M and does not move due to the frictional force.In this case the coefficient of friction is μ_s.In terms of the variables given and g,what is the magnitude of the frictional force in this case?

Accepted Answer

The answer of A sled is on a horizontal surface...

Question 8

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 at rest or moving at a constant velocity will continue to do so unless acted upon by an external net force. Since the car is moving forward at a constant velocity, it means that there is no net force acting on the tires. Therefore, the correct answer is D) zero.

Question 9

A 200-kg box is placed on a ramp.As one end of the ramp is raised,the box begins to move downward just as the angle of inclination reaches 18.0°.What is the coefficient of static friction between box and ramp?

Accepted Answer

The coefficient of static friction is equal to the tangent of the angle of inclination at which the box begins to slide. Therefore, μ = tan(18.0°) ≈ 0.32.

Question 10

A box is to be moved across a level surface.A force of magnitude 200 N may be applied at an angle of 30° below the horizontal to push the box or at an angle of 30° above the horizontal to pull the box,either application sufficient to overcome friction and move the box.Which application will cause the box to have the greater acceleration?

Accepted Answer

The force above the horizontal will have a greater horizontal component than the force below the horizontal, since the angle in the former is closer to the horizontal. The horizontal component of the force is what contributes to the acceleration of the box, and thus the force above the horizontal will result in a greater acceleration.

Question 11

​What forces are included in a free-body diagram for an object?

Accepted Answer

A free-body diagram is a visual representation of all the external forces acting on an object, therefore only the forces acting on the object need to be included. The forces the object exerts on other objects are not included in a free-body diagram for that specific object. Internal forces within the object are also not included in the diagram unless they influence the motion of the object as a whole.

Question 12

A 500-N block,on a 30.0° incline,is being held motionless by friction.The coefficient of static friction between the block and the plane is 0.63.The force due to friction is:

Accepted Answer

The answer of A 500-N block,on a 30.0° incline,is being...

Question 13

A crate of weight w is being pushed across a horizontal surface by a force directed at an angle of 20° below the horizontal.What is the magnitude of the normal force on the crate?

Accepted Answer

The normal force is the perpendicular force exerted by the surface on the crate to support its weight. This force is equal in magnitude and opposite in direction to the weight of the crate. However, since the crate is being pushed at an angle below the horizontal, there is a horizontal component of the pushing force that opposes the static friction force, which in turn reduces the normal force. In other words, the crate is being pushed downwards as well as horizontally. Therefore, the normal force is less than the weight of the crate. However, since the pushing force is at an angle less than 90°, the normal force will be more than the vertical component of the pushing force. Ultimately, the normal force is greater than the weight of the crate, so the answer is C.

Question 14

A 7.00-kg mass is placed on a 28.0° incline and friction keeps it from sliding.The coefficient of static friction in this case is 0.574,and the coefficient of sliding friction is 0.528.What is the frictional force in this situation?

Accepted Answer

The frictional force can be calculated using the formula $f = \mu_s N$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force. The normal force $N$ can be found from $N = mg \cos(\theta)$, where $m$ is the mass, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $\theta$ is the angle of the incline. Substituting the given values, $N = 7.00 \times 9.8 \times \cos(28.0^\circ) \approx 60.6 \, \text{N}$. Then, the frictional force $f = 0.574 \times 60.6 \, \text{N} \approx 34.8 \, \text{N}$. However, the correct calculation should yield a result closer to 32.2 N, indicating a mistake in the final step of the explanation. The correct calculation is $f = \mu_s N = 0.574 \times 60.6 \, \text{N} \approx 34.8 \, \text{N}$, but due to rounding or calculation error in the explanation, the intended answer is A) 32.2 N, which seems to be a mistake in the explanation process. The correct approach and calculation lead to an answer that should be around 34.8 N based on the given method, indicating a discrepancy in the provided options and explanation.

Question 15

There are nine books in a stack,each with a weight of 6 N.The coefficient of friction between all the books is 0.3 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 eight books off the bottom one?

Accepted Answer

The force required to overcome friction and start sliding is given by $F = \mu N$, where $F$ is the force of friction, $\mu$ is the coefficient of friction, and $N$ is the normal force. Here, the normal force for the top eight books is $8 \times 6\,N = 48\,N$. With a coefficient of friction of 0.3, the force required is $F = 0.3 \times 48\,N = 14.4\,N$.

Question 16

​The figure shows a block on an inclined plane. 
What is the expression for the normal force acting on the block?

Accepted Answer

The answer of ​The figure shows a block on an...

Question 17

The maximum possible value for the coefficient of static friction is:

Accepted Answer

The maximum possible value for the coefficient of static friction is greater than 1.00 when the surface of contact is not smooth and there are irregularities that can increase the friction force. This is known as the limit of static friction or the "sticking limit." However, this value is not universal and can vary depending on the materials involved and the conditions of the surface.

Question 18

A box of mass m is placed on an incline with angle of inclination θ.The box does not slide.The magnitude of the frictional force in this case is:

Accepted Answer

The frictional force must balance the component of gravitational force parallel to the incline, which is mg sin θ, to prevent the box from sliding.

Question 19

​Which is the correct expression for the force of static friction,where n is the normal force?

Accepted Answer

The expression for the force of static friction is given by f_static = μ_static*n, where μ_static is the coefficient of static friction. Option B contains the correct Unicode characters for the Greek letter mu (μ) which represents the coefficient of static friction.

Question 20

The coefficient of static friction between the tires of a car and the street is μ_s = 0.70.Of the following,what is the steepest inclination angle of a street on which a car can be parked (with wheels locked)without slipping?

Accepted Answer

The maximum angle of inclination of the street without slipping can be calculated using the equation tan(θ) = μS, where μS is the coefficient of static friction. Rearranging the equation, we get θ = tan−1(μS). Substituting the given value of μS, we get θ = tan−1(0.70) ≈ 35°. Therefore, the steepest inclination angle of a street on which a car can be parked (with wheels locked) without slipping is approximately 35°. Option C) is the correct choice.

Quiz 4: Newtons Laws of Motion

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:

A block is launched up an incline plane.After going up the plane,it slides back down to its starting position.The coefficient of friction between the block and the plane is 0.3.The speed of the block when it reaches the starting position on the trip down:

A horizontal force of 850 N is needed to overcome the force of static friction between a level floor and a 350-kg crate.If g = 9.8 m/s²,what is the coefficient of static friction?

Hector drives a pickup truck horizontally at 9.00 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 0.740,what is the minimum stopping distance for the truck so the crate will not slide?

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.

A 400-N crate rests on a ramp;the maximum angle just before it slips is 15.0° with the horizontal.What is the coefficient of static friction between crate and ramp surfaces?

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

A 200-kg box is placed on a ramp.As one end of the ramp is raised,the box begins to move downward just as the angle of inclination reaches 18.0°.What is the coefficient of static friction between box and ramp?

What forces are included in a free-body diagram for an object?

A 500-N block,on a 30.0° incline,is being held motionless by friction.The coefficient of static friction between the block and the plane is 0.63.The force due to friction is:

A crate of weight w is being pushed across a horizontal surface by a force directed at an angle of 20° below the horizontal.What is the magnitude of the normal force on the crate?

A 7.00-kg mass is placed on a 28.0° incline and friction keeps it from sliding.The coefficient of static friction in this case is 0.574,and the coefficient of sliding friction is 0.528.What is the frictional force in this situation?

There are nine books in a stack,each with a weight of 6 N.The coefficient of friction between all the books is 0.3 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 eight books off the bottom one?

The figure shows a block on an inclined plane. What is the expression for the normal force acting on the block?

The maximum possible value for the coefficient of static friction is:

A box of mass m is placed on an incline with angle of inclination θ.The box does not slide.The magnitude of the frictional force in this case is:

Which is the correct expression for the force of static friction,where n is the normal force?

The coefficient of static friction between the tires of a car and the street is μ_s = 0.70.Of the following,what is the steepest inclination angle of a street on which a car can be parked (with wheels locked) without slipping?

Quiz 4: Newtons Laws of Motion

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:

A block is launched up an incline plane.After going up the plane,it slides back down to its starting position.The coefficient of friction between the block and the plane is 0.3.The speed of the block when it reaches the starting position on the trip down:

A horizontal force of 850 N is needed to overcome the force of static friction between a level floor and a 350-kg crate.If g = 9.8 m/s2,what is the coefficient of static friction?

Hector drives a pickup truck horizontally at 9.00 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 0.740,what is the minimum stopping distance for the truck so the crate will not slide?

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 μA and μB.Choose the last answer that is correct.

A 400-N crate rests on a ramp;the maximum angle just before it slips is 15.0° with the horizontal.What is the coefficient of static friction between crate and ramp surfaces?

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

A 200-kg box is placed on a ramp.As one end of the ramp is raised,the box begins to move downward just as the angle of inclination reaches 18.0°.What is the coefficient of static friction between box and ramp?

​What forces are included in a free-body diagram for an object?

A 500-N block,on a 30.0° incline,is being held motionless by friction.The coefficient of static friction between the block and the plane is 0.63.The force due to friction is:

A crate of weight w is being pushed across a horizontal surface by a force directed at an angle of 20° below the horizontal.What is the magnitude of the normal force on the crate?

A 7.00-kg mass is placed on a 28.0° incline and friction keeps it from sliding.The coefficient of static friction in this case is 0.574,and the coefficient of sliding friction is 0.528.What is the frictional force in this situation?

There are nine books in a stack,each with a weight of 6 N.The coefficient of friction between all the books is 0.3 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 eight books off the bottom one?

​The figure shows a block on an inclined plane. What is the expression for the normal force acting on the block?

The maximum possible value for the coefficient of static friction is:

A box of mass m is placed on an incline with angle of inclination θ.The box does not slide.The magnitude of the frictional force in this case is:

​Which is the correct expression for the force of static friction,where n is the normal force?

The coefficient of static friction between the tires of a car and the street is μs = 0.70.Of the following,what is the steepest inclination angle of a street on which a car can be parked (with wheels locked) without slipping?

A horizontal force of 850 N is needed to overcome the force of static friction between a level floor and a 350-kg crate.If g = 9.8 m/s²,what is the coefficient of static friction?

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.

What forces are included in a free-body diagram for an object?

The figure shows a block on an inclined plane. What is the expression for the normal force acting on the block?

Which is the correct expression for the force of static friction,where n is the normal force?

The coefficient of static friction between the tires of a car and the street is μ_s = 0.70.Of the following,what is the steepest inclination angle of a street on which a car can be parked (with wheels locked) without slipping?