Question 1

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 force pushing the crate has two components, one perpendicular to the surface and one parallel to the surface. The perpendicular component balances out the normal force, so the magnitude of the normal force is equal to the weight of the crate, which is W. The force pushing the crate also has a horizontal component, which is opposed by friction. However, the magnitude of the normal force is not affected by the coefficient of kinetic friction, so the answer is not affected by the information that the coefficient is not given. Therefore, choice C is the correct answer.

Question 2

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 3

Two blocks are at rest on a horizontal frictionless surface with a compressed spring between them.The 5.0-kg block accelerates when the string holding the system in compression is cut and the 10-kg box accelerates in the opposite direction.What is the magnitude of the force from the expanding spring on the 10-kg box when the 5.0-kg box is undergoing an acceleration of 9.0 m/s²?

Accepted Answer

The force on the 5.0-kg block is F = ma = 5.0 kg * 9.0 m/s² = 45 N. By Newton's third law, the force on the 10-kg block is equal in magnitude and opposite in direction, so it is also 45 N.

Question 4

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 way down will be less than the launching speed. This is because the frictional force between the block and the incline plane will slow it down on the way up and it will continue to act on the block on its way down, opposing its motion and reducing its speed. The coefficient of friction being 0.3 tells us the magnitude of the frictional force, but we don't have any other information about the forces acting on the block, so we cannot compare the launch speed and the return speed directly.

Question 5

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 6

The net force on an object is in the positive x-direction.Consider the following statements.
(I.
The object can be moving in the negative x-direction.
(II.
The object can be speeding up.
(III)
The object can be slowing down.
(IV.
The object can be moving in the positive y-direction.
Which of the statements are true?

Accepted Answer

All the statements are true. A net force in the positive x-direction means the acceleration of the object is in the positive x-direction. (I) An object can be moving in the negative x-direction but slowing down due to the force in the opposite direction. (II) The object can be speeding up if it was already moving in the positive x-direction or if it was moving in the negative x-direction and the force has now overcome its velocity, causing it to accelerate. (III) An object moving in the negative x-direction can be slowing down if the force in the positive x-direction is reducing its negative velocity. (IV) The object can also be moving in the positive y-direction, as the force in the x-direction does not preclude motion in the y-direction; the two are orthogonal and can occur simultaneously.

Question 7

The coefficient of static friction between the tires of a car and the street is μ_s = 0.67.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 angle θ at which a car can be parked without slipping is given by θ = arctan(μ_s). For μ_s = 0.67, θ = arctan(0.67) ≈ 34°.

Question 8

An object weighs 300 N.If the gravitational constant G were half of what it is currently,what would the weight of the object be?

Accepted Answer

Weight, W = m*g, where m is mass and g is gravitational acceleration. So, if G is half, then the gravitational force on the object is half of what it would be normally. Hence, weight of the object = (1/2)*m*(g/2) = (1/4)*m*g. Therefore, the weight would be halved, i.e., 150 N. So, the correct choice is B.

Question 9

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 10

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

Accepted Answer

The coefficient of static friction is defined as the ratio of the maximum force of static friction to the normal force. As the maximum force of static friction is dependent on the actual weight of the object, there is no theoretical limit to the coefficient of static friction. Therefore, answer D, "greater than 1.00," is the best choice. However, in reality, the coefficient of static friction is limited by the materials and surface conditions of the objects in contact, so it is highly unlikely to have a coefficient of static friction greater than 1.00 in real-world situations.

Question 11

A 150-N block,on a 25° incline,is being held motionless by friction.The coefficient of static friction between the block and the plane is 0.52.The force due to friction is:

Accepted Answer

First, draw a diagram of the situation to help visualize the forces acting on the block.

The force of gravity (weight) can be split into components parallel and perpendicular to the incline.

F_gravity_parallel = m*g*sin(theta) 
= 150 N * sin(25°) 
= 63 N

This is the force that would cause the block to slide down the incline if there were no friction.

The force of static friction must be equal and opposite to this force in order to keep the block from moving.

F_friction = F_gravity_parallel 
= 63 N

Using the given coefficient of friction:

F_friction = friction_coefficient * F_normal 
where F_normal is the normal force acting perpendicular to the incline.

Since the block is not accelerating vertically (it's being held motionless), the normal force must be equal and opposite to the perpendicular component of the weight:

F_normal = m*g*cos(theta) 
= 150 N * cos(25°) 
= 136 N

Plugging this back into the friction equation:

F_friction = friction_coefficient * F_normal 
= 0.52 * 136 N 
= 70.72 N

This is larger than the force required to keep the block motionless (63 N), which means the block is not actually held in place by the maximum possible static friction force. The coefficient of friction tells us how large the friction force can be before the block starts to slide, not how large it must be to keep the block stationary. Therefore, the force due to friction is 63 N, which is Answer C.

Question 12

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

Question 13

Five boxes,each having weight 150 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 150 N, the total weight of the five boxes is:

Total weight = 150 N/box x 5 boxes = 750 N

By Newton's third law of motion, the normal force upward exerted by the fourth box from the top on the third box from the top is equal in magnitude to the normal force downward exerted by the third box on the fourth box.

Therefore, the normal force between the third and fourth boxes is half of the weight of the three boxes above them:

Normal force between the third and fourth boxes = 750 N/2 = 375 N

However, we need to find the normal force exerted by the fourth box on the third box, not vice versa. Since the boxes are at rest, these normal forces are equal in magnitude and opposite in direction.

Therefore, the magnitude of the normal force upward exerted by the fourth box from the top on the third box from the top is also 375 N/2 = 187.5 N, which is closest to C, 450 N.

Quiz 4: The Laws of Motion

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 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.

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:

The coefficient of static friction between the tires of a car and the street is μ_s = 0.67.Of the following,what is the steepest inclination angle of a street on which a car can be parked (with wheels locked) without slipping?

An object weighs 300 N.If the gravitational constant G were half of what it is currently,what would the weight of the object be?

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:

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

A 150-N block,on a 25° incline,is being held motionless by friction.The coefficient of static friction between the block and the plane is 0.52.The force due to 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:

Five boxes,each having weight 150 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?

Quiz 4: The Laws of Motion

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 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.

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:

The coefficient of static friction between the tires of a car and the street is μs = 0.67.Of the following,what is the steepest inclination angle of a street on which a car can be parked (with wheels locked) without slipping?

An object weighs 300 N.If the gravitational constant G were half of what it is currently,what would the weight of the object be?

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:

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

A 150-N block,on a 25° incline,is being held motionless by friction.The coefficient of static friction between the block and the plane is 0.52.The force due to 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:

Five boxes,each having weight 150 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 μ_s = 0.67.Of the following,what is the steepest inclination angle of a street on which a car can be parked (with wheels locked) without slipping?