Question 1

A crate is in the bed of a pickup truck on a level road.The coefficients of static friction and kinetic friction between the crate and the bed of the truck are  and  ,respectively.When the truck starts from rest,what is the limiting acceleration that it can have before the crate starts sliding along the truck bed?

Accepted Answer

The limiting acceleration is given by the coefficient of static friction times the gravitational acceleration. Therefore, the limiting acceleration is:

(11eaa64f_8f68_aada_a696_4744cf2170b6_TB5258_11) * g = (11eaa64f_8f68_f8fe_a696_2fdb62d2940e_TB5258_11) * 9.8 m/s^2

Simplifying, we get:

a = 107.78 m/s^2

Rounding to two significant figures, we get:

a = 1.1 * 10^2 m/s^2 = 11eaa64f_8f68_f8fe_a696_2fdb62d2940e_TB5258_11

Question 2

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 object in question is being held in place by adhesion rather than simply surface roughness. This can occur with materials such as rubber or certain types of tape, where the molecules of the object and the surface it is on actually bond together. However, it is important to note that this is not a common occurrence and most surfaces have a coefficient of static friction that is less than or equal to 1.00.

Question 3

A 9.0-kg hanging weight is connected by a string over a pulley to a 5.0-kg block sliding on a flat table.If the coefficient of sliding friction is 0.20,find the tension in the string.

Accepted Answer

The answer of A 9.0-kg hanging weight is connected by...

Question 4

A box is to be moved across a level surface.A force of magnitude 200 N may be applied at an angle of 30 $\degree$below the horizontal to push the box or at an angle of 30 $\degree$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 component in the direction of motion than the force below the horizontal. This can be seen as follows:
For the force above the horizontal, the component in the direction of motion is given by:
Fcos30° = 200cos30° = 173.2 N
For the force below the horizontal, the component in the direction of motion is given by:
Fcos(180°-30°) = Fcos150° = 200cos150° = -173.2 N (since cos150° = -cos30°)
Therefore, the force above the horizontal will cause the box to have a greater acceleration.

Question 5

An object weighs 100 N.If the gravitational constant G were half of what it is currently,what would the weight of the object be?

Accepted Answer

Weight is calculated by multiplying the mass of the object by gravitational acceleration g, which depends on the value of G. So, if G is decreased to half, the value of g will also decrease to half. Therefore, the weight of the object will be half of its original weight (100 N/2 = 50 N).

Question 6

The net force on an object is in the positive x-direction.Consider the following statements.  Which of the statements are true?

Accepted Answer

The answer of The net force on an object is...

Question 7

Two blocks are at rest on a horizontal frictionless surface with a compressed spring between them.The 10-kg block accelerates when the string holding the system in compression is cut and the 20-kg box accelerates in the opposite direction.What is the magnitude of the force from the expanding spring on the 20-kg box when the 10-kg box is undergoing an acceleration of 8.0  ?

Accepted Answer

We can use the conservation of momentum to find the magnitude of the velocity of each block after the spring is released:

m1v1 = m2v2

10kg (0 m/s) = 20 kg (v2)

v2 = 0 m/s

This means that the 20-kg block does not move initially and only starts moving because of the force from the expanding spring.

Next, we can use Newton's second law to find the magnitude of the force on the 10-kg block:

F = ma

F = 10 kg x 8.0 m/s^2

F = 80 N

Since the spring is compressed between the blocks, the force on the 10-kg block is equal and opposite to the force on the 20-kg block. Therefore, the magnitude of the force from the expanding spring on the 20-kg box is also 80 N, which corresponds to choice D.

Question 8

Five boxes,each having weight 100 N,are stacked up.The bottom box is the fifth from the top.What is the magnitude of the normal force upward exerted by the fourth box from the top on the third box from the top?

Accepted Answer

Since each box has a weight of 100 N, the total weight of the five boxes is 500 N. When they are stacked up, the bottom box experiences a normal force of 500 N from the ground. This normal force is transmitted through the boxes above it, resulting in a decreasing normal force as we move up the stack.

Therefore, the fourth box from the top is experiencing a normal force of 200 N (from the box directly below it) and is exerting a downward force of 200 N on the third box from the top. However, the third box from the top is also experiencing a normal force of 100 N from the box directly below it.

Thus, the net normal force exerted by the fourth box from the top on the third box from the top is 200 N - 100 N = 100 N.

Therefore, the correct answer is:

C) 300 N

Question 9

Two boxes are stacked on top of one another on the back of a flat bed truck on a level roadway.The coefficient of static friction between the boxes is  ,and the coefficient of static friction between the bottom box and the bed of the truck is  .What is the limiting acceleration the truck can undergo before one or both of the boxes start sliding if  ?

Accepted Answer

The limiting acceleration can be found by using the formula:
a = μg
where μ is the coefficient of static friction and g is the acceleration due to gravity (9.8 m/s^2).
For the top box to start sliding:
a < μtop g
For the bottom box to start sliding:
a < μbottom g
The limiting acceleration will be the minimum of the above two values.
Given:
μtop = 0.11
μbottom = 0.116
g = 9.8 m/s^2

Substituting the given values in the above formula:
For the top box to start sliding: 
a < 0.11 × 9.8 = 1.078 m/s^2
For the bottom box to start sliding:
a < 0.116 × 9.8 = 1.137 m/s^2
The limiting acceleration will be the minimum of the above two values:
a < 1.078 m/s^2
Thus, the limiting acceleration of the truck is 1.078 m/s^2. 
Therefore, the answer is C.

Question 10

The coefficient of static friction between the tires of a car and the street is m_s = 0.77.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 steepest inclination angle can be found using the formula:
tan θ = μs
where μs is the coefficient of static friction.
Substituting the given value of μs = 0.77, we get:
tan θ = 0.77
Taking inverse tangent on both sides, we get:
θ = tan⁻¹0.77
θ ≈ 37° (rounded to the nearest degree)
Therefore, the steepest inclination angle of a street on which a car can be parked without slipping is 37°, which is option C.

Question 11

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 block experiences frictional force while going up the incline, which reduces its speed. When it slides back down, it will also experience frictional force in the opposite direction, further reducing its speed. Therefore, the speed of the block when it reaches the starting position on the trip down will be less than the launching speed. The exact speed cannot be calculated without additional information such as the height of the incline and the initial speed of the block.

Question 12

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 time for the trip up the plane:

Accepted Answer

The answer of A block is launched up an incline...

Question 13

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

Accepted Answer

The normal force is more than W because the vertical component of the applied force adds to the weight of the crate, increasing the normal force.

Question 14

A 100-N block,on a 30 $\degree$incline,is being held motionless by friction.The coefficient of static friction between the block and the plane is 0.60.The force due to friction is:

Accepted Answer

The force due to friction is given by the product of the coefficient of static friction and the normal force. The normal force is the component of the weight of the block perpendicular to the plane, which is given by N = mg cos(theta), where m is the mass of the block, g is the acceleration due to gravity, and theta is the angle of the incline. 
In this case, m = 100 N / 9.81 m/s^2 = 10.18 kg, theta = 30 degrees, and cos(theta) = sqrt(3)/2. Therefore, N = 10.18 kg x 9.81 m/s^2 x sqrt(3)/2 = 50 N. 
So, the force due to friction is 0.6 x 50 N = 30 N, which is the closest value to the options given. Therefore, the answer is choice C, 50 N.

Question 15

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

Accepted Answer

The answer of A box of mass m is placed...

Quiz 4: The Laws of Motion

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

A 9.0-kg hanging weight is connected by a string over a pulley to a 5.0-kg block sliding on a flat table.If the coefficient of sliding friction is 0.20,find the tension in the string.

An object weighs 100 N.If the gravitational constant G were half of what it is currently,what would the weight of the object be?

The net force on an object is in the positive x-direction.Consider the following statements. Which of the statements are true?

Five boxes,each having weight 100 N,are stacked up.The bottom box is the fifth from the top.What is the magnitude of the normal force upward exerted by the fourth box from the top on the third box from the top?

The coefficient of static friction between the tires of a car and the street is m_s = 0.77.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 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 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 time for the trip up the plane:

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

A 100-N block,on a 30 $\degree$ incline,is being held motionless by friction.The coefficient of static friction between the block and the plane is 0.60.The force due to friction is:

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

Quiz 4: The Laws of Motion

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

A 9.0-kg hanging weight is connected by a string over a pulley to a 5.0-kg block sliding on a flat table.If the coefficient of sliding friction is 0.20,find the tension in the string.

An object weighs 100 N.If the gravitational constant G were half of what it is currently,what would the weight of the object be?

The net force on an object is in the positive x-direction.Consider the following statements. Which of the statements are true?

Five boxes,each having weight 100 N,are stacked up.The bottom box is the fifth from the top.What is the magnitude of the normal force upward exerted by the fourth box from the top on the third box from the top?

The coefficient of static friction between the tires of a car and the street is ms = 0.77.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 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 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 time for the trip up the plane:

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

A 100-N block,on a 30 °\degree° incline,is being held motionless by friction.The coefficient of static friction between the block and the plane is 0.60.The force due to friction is:

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

The coefficient of static friction between the tires of a car and the street is m_s = 0.77.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 crate of weight W is being pushed across a horizontal surface by a force directed at an angle of 20 $\degree$ below the horizontal.What is the magnitude of the normal force on the crate?

A 100-N block,on a 30 $\degree$ incline,is being held motionless by friction.The coefficient of static friction between the block and the plane is 0.60.The force due to friction is: