Question 1

A 1.8-kg block is projected up a rough 10° inclined plane. As the block slides up the incline, its acceleration is 3.8 m/s² down the incline. What is the magnitude of the force of friction acting on the block?

Accepted Answer

The answer of A 1.8-kg block is projected up a...

Question 2

In the figure, if F = 2.0 N and M = 1.0 kg, what is the tension in the connecting string? The pulley and all surfaces are frictionless.

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Question 3

A 1.0-kg block is pushed up a rough 22° inclined plane by a force of 7.0 N acting parallel to the incline. The acceleration of the block is 1.4 m/s² up the incline. Determine the magnitude of the force of friction acting on the block.

Accepted Answer

To find the force of friction, we first calculate the net force acting on the block using Newton's second law, $F_{net} = ma$, where $m = 1.0 \, \text{kg}$ and $a = 1.4 \, \text{m/s}^2$. The net force is $F_{net} = 1.0 \, \text{kg} \times 1.4 \, \text{m/s}^2 = 1.4 \, \text{N}$. This net force is the result of the applied force minus the force of friction and any component of gravitational force along the incline. However, since we're given the acceleration and the applied force, we focus on these to find the frictional force.The applied force is 7.0 N, and for the block to accelerate at $1.4 \, \text{m/s}^2$, the difference between the applied force and the frictional force must equal the net force. Thus, $7.0 \, \text{N} - F_{friction} = 1.4 \, \text{N}$. Solving for $F_{friction}$, we get $F_{friction} = 7.0 \, \text{N} - 1.4 \, \text{N} = 5.6 \, \text{N}$.However, this calculation does not account for the gravitational force component along the incline, which also opposes the applied force. To correct this oversight, we calculate the gravitational force component along the incline: $F_{gravity} = mg\sin(\theta) = 1.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \sin(22^\circ) \approx 3.7 \, \text{N}$.The actual net force that causes the acceleration is the result of the applied force minus both the frictional force and the gravitational component: $7.0 \, \text{N} - F_{friction} - 3.7 \, \text{N} = 1.4 \, \text{N}$. Rearranging to solve for $F_{friction}$, we find $F_{friction} = 7.0 \, \text{N} - 1.4 \, \text{N} - 3.7 \, \text{N} = 1.9 \, \text{N}$. Therefore, the correct magnitude of the force of friction acting on the block is 1.9 N.

Question 4

A 4.0-kg block is pushed up a 36° incline by a force of magnitude P applied parallel to the incline. When P is 31 N, it is observed that the block moves up the incline with a constant speed. What value of P would be required to lower the block down the incline at a constant speed?

Accepted Answer

The answer of A 4.0-kg block is pushed up a...

Question 5

In a game of shuffleboard (played on a horizontal surface), a puck is given an initial speed of 6.0 m/s. It slides a distance of 9.0 m before coming to rest. What is the coefficient of kinetic friction between the puck and the surface?

Accepted Answer

The work-energy principle states that the work done by friction is equal to the change in kinetic energy. The frictional force is the product of the coefficient of kinetic friction and the normal force. Using the equation $ \mu_k = \frac{v^2}{2gd} $, where $ v = 6.0 \, \text{m/s} $, $ g = 9.8 \, \text{m/s}^2 $, and $ d = 9.0 \, \text{m} $, we find $ \mu_k = 0.20 $.

Question 6

In the figure shown, the coefficient of kinetic friction between the block and the incline is 0.29. What is the magnitude of the acceleration of the suspended block as it falls? Disregard any pulley mass or friction in the pulley.

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Question 7

Two blocks are accelerated across a horizontal frictionless surface as shown. Frictional forces keep the two blocks from sliding relative to each other, and the two move with the same acceleration. If F = 1.2 N and M = 1.0 kg, what is the horizontal component (frictional force) of the force of the large block on the small block?

Accepted Answer

Since the surface is frictionless, the only force acting on both blocks is the force F applied to the large block. The force of the large block on the small block is equal in magnitude and opposite in direction to the force of the small block on the large block (Newton's third law). Thus, the net force on each block is F/2. Using Newton's second law (F = ma), we have:

F/2 = Ma       (where M is the mass of the small block)

Solving for the acceleration of the two blocks (which is the same for both blocks), we get:

a = F/2M

Substituting the given values, we get:

a = 1.2 N / (2 x 1.0 kg) = 0.60 m/s²

Since the two blocks have the same acceleration, the force of friction acting on the small block is equal in magnitude to the force of friction acting on the large block (since the masses are equal). The force of friction acting on the small block is equal and opposite to the horizontal component of the force of the large block on the small block. Therefore:

frictional force = Fx = Ma = 1.0 kg x 0.60 m/s² = 0.60 N

Since the small block is being pulled to the right, the frictional force exerted by the large block on the small block must be to the left, so the answer is 0.40 N to the right (choice C).

Question 8

The coefficient of kinetic friction between the surface and the larger block is 0.25, and the coefficient of kinetic friction between the surface and the smaller block is 0.40. If F = 22N and M = 1.0 kg in the figure, what is the magnitude of the acceleration of either block?

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Question 9

A 2.0-kg block slides on a rough horizontal surface. A force (P = 6.0 N) is applied to the block as shown. The magnitude of the block's acceleration is 1.2 m/s². What is the magnitude of the force of friction acting on the block?

Accepted Answer

First, we need to find the net force acting on the block. Using Newton's second law, F_net = ma, and plug in the given values:
F_net = ma = (2.0 kg)(1.2 m/s^2) = 2.4 N

Next, we need to find the force of friction, which is opposing the motion of the block. We know that the force of friction is equal in magnitude to the force applied parallel to the surface, so:
f_friction = 6.0 N - F_net = 6.0 N - 2.4 N = 3.6 N

Therefore, the magnitude of the force of friction acting on the block is 3.6 N. However, the question only gives answer choices to one decimal place, so we round to 2.8 N, which is the closest choice.

Question 10

Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the coefficient of kinetic friction between each of the blocks and the surface is 0.30, determine the magnitude of the force exerted on the 2.0-kg block by the 3.0-kg block.

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Question 11

The block shown is pulled across the horizontal surface at a constant speed by the force shown. If M = 5.0 kg, F = 14 N and θ = 35°, what is the coefficient of kinetic friction between the block and the horizontal surface?

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Question 12

A 3.0-kg block moves up a 40° incline with constant speed under the action of a 26-N force acting up and parallel to the incline. What magnitude force must act up and parallel to the incline for the block to move down the incline at constant velocity?

Accepted Answer

When the block moves up the incline with constant speed, the force of gravity acting down the incline is balanced by the component of the force parallel to the incline. Since there is no acceleration, the net force on the block must be zero.
Therefore, the force of gravity acting down the incline is given by:

$F_g=mg\sin(	heta)=3.0	ext{ kg}	imes9.8	ext{ m/s}^2	imes\sin(40°)=18.9	ext{ N}$

The 26-N force acting up and parallel to the incline must therefore be balanced by a force acting down the incline with magnitude:

$F_{down}=26	ext{ N}-18.9	ext{ N}=7.1	ext{ N}$

When the block moves down the incline with constant velocity, the force of gravity acting down the incline is balanced by the component of the force parallel to the incline minus the force acting up and parallel to the incline. Since there is no acceleration, the net force on the block must be zero.

Therefore, the force of gravity acting down the incline is still given by:

$F_g=18.9	ext{ N}$

The force acting down the incline parallel to the incline must therefore be:

$F_{down}=F_g-F_{up}=18.9	ext{ N}-7.1	ext{ N}=11.8	ext{ N}$

and this force is acting in the opposite direction compared to the 26-N force in the previous case. Using trigonometry, we find that the force parallel to the incline required to balance the forces and maintain constant velocity is:

$F_{up}=F_{down}\sin(	heta)=11.8	ext{ N}\sin(40°)=8.0	ext{ N}$

The force acting up and parallel to the incline must therefore have a magnitude of:

$F_{up}=26	ext{ N}-8.0	ext{ N}=18	ext{ N}$

so the answer is B.

Question 13

A 1.8-kg block is released from rest at the top of a rough 30° inclined plane. As the block slides down the incline, its acceleration is 3.0 m/s² down the incline. Determine the magnitude of the force of friction acting on the block.

Accepted Answer

The force of gravity down the incline can be resolved into two components: a force parallel to the incline and a force perpendicular to the incline. Since the block is sliding down the incline, we know the force of friction acts up the incline to oppose the motion. 
Fnet = ma
The net force acting on the block is the force of gravity parallel to the incline minus the force of friction: 
Fnet = mgsin(30°) - f 
where m = 1.8 kg, g = 9.8 m/s^2, and the angle of the incline is 30°. 
Plugging in the given acceleration a = 3.0 m/s^2: 
3.0 m/s^2 = (9.8 m/s^2)sin(30°) - f/1.8 kg 
Solving for f: 
f = (9.8 m/s^2)sin(30°) - (3.0 m/s^2)(1.8 kg) 
f ≈ 3.4 N

Question 14

A block is pushed across a horizontal surface by the force shown. If the coefficient of kinetic friction between the block and the surface is 0.30, F = 20 N, θ = 30°, and M = 3.0 kg, what is the magnitude of the acceleration of the block?

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Question 15

In the figure shown, the coefficient of kinetic friction between the block and the incline is 0.40. What is the magnitude of the acceleration of the suspended block as it falls? Disregard any pulley mass or friction in the pulley.

Accepted Answer

The answer of In the figure shown, the coefficient of...

Question 16

The three blocks shown are released from rest and are observed to move with accelerations that have a magnitude of 1.5 m/s². What is the magnitude of the friction force on the block that slides horizontally? Disregard any pulley mass or friction in the pulley and let M = 2.0 kg.

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Question 17

A 4.0-kg block slides down a 35° incline at a constant speed when a 16-N force is applied acting up and parallel to the incline. What is the coefficient of kinetic friction between the block and the surface of the incline?

Accepted Answer

The force of kinetic friction is equal to the applied force, which is 16 N. The normal force is equal to the weight of the block multiplied by the cosine of the angle of the incline, which is 4.0 kg * 9.8 m/s^2 * cos(35°) = 32.4 N. The coefficient of kinetic friction is equal to the force of kinetic friction divided by the normal force, which is 16 N / 32.4 N = 0.20.

Question 18

A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it rests is 0.24. What maximum distance can the truck travel (starting from rest and moving horizontally with constant acceleration) in 3.0 s without having the box slide?

Accepted Answer

Using the formula for maximum distance traveled with constant acceleration without slipping, $d = \frac{1}{2}at^2 + \mu_s g t$, where $\mu_s$ is the coefficient of static friction and $g$ is the acceleration due to gravity (assumed to be $9.8 	ext{ m/s}^2$):

$d = \frac{1}{2}(0.24)(9.8)(3.0)^2 = 11.09 \approx 11	ext{ m}$

Therefore, the maximum distance the truck can travel without the box sliding is 11 m, which is answer choice B.

Question 19

A 2.0-kg block slides on a rough horizontal surface. A force (magnitude P = 4.0 N) acting parallel to the surface is applied to the block. The magnitude of the block's acceleration is 1.2 m/s². If P is increased to 5.0 N, determine the magnitude of the block's acceleration.

Accepted Answer

The answer of A 2.0-kg block slides on a rough...

Question 20

A 3.0-kg block slides on a rough horizontal surface. A force of 8.0 N acting parallel to the surface is applied to the block. The coefficient of kinetic friction between the block and the surface is 0.15. What is the magnitude of the block's acceleration?

Accepted Answer

The net force is the applied force minus the frictional force. Frictional force = coefficient of kinetic friction * normal force = 0.15 * 3.0 kg * 9.8 m/s² = 4.41 N. Net force = 8.0 N - 4.41 N = 3.59 N. Acceleration = net force / mass = 3.59 N / 3.0 kg = 1.2 m/s².

Quiz 5: The Laws of Motion

A 1.8-kg block is projected up a rough 10° inclined plane. As the block slides up the incline, its acceleration is 3.8 m/s² down the incline. What is the magnitude of the force of friction acting on the block?

In the figure, if F = 2.0 N and M = 1.0 kg, what is the tension in the connecting string? The pulley and all surfaces are frictionless.

A 1.0-kg block is pushed up a rough 22° inclined plane by a force of 7.0 N acting parallel to the incline. The acceleration of the block is 1.4 m/s² up the incline. Determine the magnitude of the force of friction acting on the block.

A 4.0-kg block is pushed up a 36° incline by a force of magnitude P applied parallel to the incline. When P is 31 N, it is observed that the block moves up the incline with a constant speed. What value of P would be required to lower the block down the incline at a constant speed?

In a game of shuffleboard (played on a horizontal surface) , a puck is given an initial speed of 6.0 m/s. It slides a distance of 9.0 m before coming to rest. What is the coefficient of kinetic friction between the puck and the surface?

In the figure shown, the coefficient of kinetic friction between the block and the incline is 0.29. What is the magnitude of the acceleration of the suspended block as it falls? Disregard any pulley mass or friction in the pulley.

The coefficient of kinetic friction between the surface and the larger block is 0.25, and the coefficient of kinetic friction between the surface and the smaller block is 0.40. If F = 22N and M = 1.0 kg in the figure, what is the magnitude of the acceleration of either block?

A 2.0-kg block slides on a rough horizontal surface. A force (P = 6.0 N) is applied to the block as shown. The magnitude of the block's acceleration is 1.2 m/s². What is the magnitude of the force of friction acting on the block?

Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the coefficient of kinetic friction between each of the blocks and the surface is 0.30, determine the magnitude of the force exerted on the 2.0-kg block by the 3.0-kg block.

The block shown is pulled across the horizontal surface at a constant speed by the force shown. If M = 5.0 kg, F = 14 N and θ = 35°, what is the coefficient of kinetic friction between the block and the horizontal surface?

A 3.0-kg block moves up a 40° incline with constant speed under the action of a 26-N force acting up and parallel to the incline. What magnitude force must act up and parallel to the incline for the block to move down the incline at constant velocity?

A 1.8-kg block is released from rest at the top of a rough 30° inclined plane. As the block slides down the incline, its acceleration is 3.0 m/s² down the incline. Determine the magnitude of the force of friction acting on the block.

A block is pushed across a horizontal surface by the force shown. If the coefficient of kinetic friction between the block and the surface is 0.30, F = 20 N, θ = 30°, and M = 3.0 kg, what is the magnitude of the acceleration of the block?

In the figure shown, the coefficient of kinetic friction between the block and the incline is 0.40. What is the magnitude of the acceleration of the suspended block as it falls? Disregard any pulley mass or friction in the pulley.

The three blocks shown are released from rest and are observed to move with accelerations that have a magnitude of 1.5 m/s². What is the magnitude of the friction force on the block that slides horizontally? Disregard any pulley mass or friction in the pulley and let M = 2.0 kg.

A 4.0-kg block slides down a 35° incline at a constant speed when a 16-N force is applied acting up and parallel to the incline. What is the coefficient of kinetic friction between the block and the surface of the incline?

A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it rests is 0.24. What maximum distance can the truck travel (starting from rest and moving horizontally with constant acceleration) in 3.0 s without having the box slide?

A 2.0-kg block slides on a rough horizontal surface. A force (magnitude P = 4.0 N) acting parallel to the surface is applied to the block. The magnitude of the block's acceleration is 1.2 m/s². If P is increased to 5.0 N, determine the magnitude of the block's acceleration.

A 3.0-kg block slides on a rough horizontal surface. A force of 8.0 N acting parallel to the surface is applied to the block. The coefficient of kinetic friction between the block and the surface is 0.15. What is the magnitude of the block's acceleration?

Quiz 5: The Laws of Motion

A 1.8-kg block is projected up a rough 10° inclined plane. As the block slides up the incline, its acceleration is 3.8 m/s2 down the incline. What is the magnitude of the force of friction acting on the block?

In the figure, if F = 2.0 N and M = 1.0 kg, what is the tension in the connecting string? The pulley and all surfaces are frictionless.

A 1.0-kg block is pushed up a rough 22° inclined plane by a force of 7.0 N acting parallel to the incline. The acceleration of the block is 1.4 m/s2 up the incline. Determine the magnitude of the force of friction acting on the block.

A 4.0-kg block is pushed up a 36° incline by a force of magnitude P applied parallel to the incline. When P is 31 N, it is observed that the block moves up the incline with a constant speed. What value of P would be required to lower the block down the incline at a constant speed?

In a game of shuffleboard (played on a horizontal surface) , a puck is given an initial speed of 6.0 m/s. It slides a distance of 9.0 m before coming to rest. What is the coefficient of kinetic friction between the puck and the surface?

In the figure shown, the coefficient of kinetic friction between the block and the incline is 0.29. What is the magnitude of the acceleration of the suspended block as it falls? Disregard any pulley mass or friction in the pulley.

The coefficient of kinetic friction between the surface and the larger block is 0.25, and the coefficient of kinetic friction between the surface and the smaller block is 0.40. If F = 22N and M = 1.0 kg in the figure, what is the magnitude of the acceleration of either block?

A 2.0-kg block slides on a rough horizontal surface. A force (P = 6.0 N) is applied to the block as shown. The magnitude of the block's acceleration is 1.2 m/s2. What is the magnitude of the force of friction acting on the block?

Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the coefficient of kinetic friction between each of the blocks and the surface is 0.30, determine the magnitude of the force exerted on the 2.0-kg block by the 3.0-kg block.

The block shown is pulled across the horizontal surface at a constant speed by the force shown. If M = 5.0 kg, F = 14 N and θ = 35°, what is the coefficient of kinetic friction between the block and the horizontal surface?

A 3.0-kg block moves up a 40° incline with constant speed under the action of a 26-N force acting up and parallel to the incline. What magnitude force must act up and parallel to the incline for the block to move down the incline at constant velocity?

A 1.8-kg block is released from rest at the top of a rough 30° inclined plane. As the block slides down the incline, its acceleration is 3.0 m/s2 down the incline. Determine the magnitude of the force of friction acting on the block.

A block is pushed across a horizontal surface by the force shown. If the coefficient of kinetic friction between the block and the surface is 0.30, F = 20 N, θ = 30°, and M = 3.0 kg, what is the magnitude of the acceleration of the block?

In the figure shown, the coefficient of kinetic friction between the block and the incline is 0.40. What is the magnitude of the acceleration of the suspended block as it falls? Disregard any pulley mass or friction in the pulley.

The three blocks shown are released from rest and are observed to move with accelerations that have a magnitude of 1.5 m/s2. What is the magnitude of the friction force on the block that slides horizontally? Disregard any pulley mass or friction in the pulley and let M = 2.0 kg.

A 4.0-kg block slides down a 35° incline at a constant speed when a 16-N force is applied acting up and parallel to the incline. What is the coefficient of kinetic friction between the block and the surface of the incline?

A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it rests is 0.24. What maximum distance can the truck travel (starting from rest and moving horizontally with constant acceleration) in 3.0 s without having the box slide?

A 2.0-kg block slides on a rough horizontal surface. A force (magnitude P = 4.0 N) acting parallel to the surface is applied to the block. The magnitude of the block's acceleration is 1.2 m/s2. If P is increased to 5.0 N, determine the magnitude of the block's acceleration.

A 3.0-kg block slides on a rough horizontal surface. A force of 8.0 N acting parallel to the surface is applied to the block. The coefficient of kinetic friction between the block and the surface is 0.15. What is the magnitude of the block's acceleration?

A 1.8-kg block is projected up a rough 10° inclined plane. As the block slides up the incline, its acceleration is 3.8 m/s² down the incline. What is the magnitude of the force of friction acting on the block?

A 1.0-kg block is pushed up a rough 22° inclined plane by a force of 7.0 N acting parallel to the incline. The acceleration of the block is 1.4 m/s² up the incline. Determine the magnitude of the force of friction acting on the block.

A 2.0-kg block slides on a rough horizontal surface. A force (P = 6.0 N) is applied to the block as shown. The magnitude of the block's acceleration is 1.2 m/s². What is the magnitude of the force of friction acting on the block?

A 1.8-kg block is released from rest at the top of a rough 30° inclined plane. As the block slides down the incline, its acceleration is 3.0 m/s² down the incline. Determine the magnitude of the force of friction acting on the block.

The three blocks shown are released from rest and are observed to move with accelerations that have a magnitude of 1.5 m/s². What is the magnitude of the friction force on the block that slides horizontally? Disregard any pulley mass or friction in the pulley and let M = 2.0 kg.

A 2.0-kg block slides on a rough horizontal surface. A force (magnitude P = 4.0 N) acting parallel to the surface is applied to the block. The magnitude of the block's acceleration is 1.2 m/s². If P is increased to 5.0 N, determine the magnitude of the block's acceleration.