Question 1

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? 
A)0.40 N to the left
B)0.80 N to the right
C)0.40 N to the right
D)0.80 N to the left
E)1.20 N to the left

Accepted Answer

Since both blocks are moving with the same acceleration, the force of friction between them must be equal in magnitude to the force exerted by the larger block on the smaller block. Using Newton's second law, F=ma, we can find the acceleration of both blocks: a=F/M=1.2 N/1.0 kg=1.2 m/s^2. Therefore, the force of friction between the blocks is Ff=ma=1.0 kg x 1.2 m/s^2=1.2 N. Since the frictional force keeps the blocks from sliding relative to each other, it must be directed to the right, opposite to the direction of the applied force. Therefore, the horizontal component of the force of the larger block on the smaller block is also to the right, equal in magnitude to the force of friction: 0.4 N (since the force is split between the two blocks, 0.8 N is the total force of the larger block on the system). The correct answer is therefore choice C.

Question 2

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)1.9 m/s² B)1.2 m/s² C)2.3 m/s² D)1.5 m/s² E)2.9 m/s²

Accepted Answer

The answer of A 3.0-kg block slides on a rough...

Question 3

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?
A)0.20
B)0.18
C)0.15
D)0.13
E)0.27

Accepted Answer

The answer of In a game of shuffleboard (played on...

Question 4

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)5.0 N B)3.8 N C)4.2 N D)4.6 N E)6.5 N

Accepted Answer

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

Question 5

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. A)3.4 m/s² B)3.7 m/s² C)4.2 m/s² D)3.9 m/s² E)5.4 m/s²

Accepted Answer

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

Question 6

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)4.2 N B)3.0 N C)3.4 N D)3.8 N E)2.3 N

Accepted Answer

First, find the component of gravity parallel to the incline:
$F_g\parallel = mg\sin	heta = (1.8	ext{ kg})(9.8	ext{ m/s}^2)(\sin30^\circ) = 8.82	ext{ N}$
Then find the net force parallel to the incline:
$F_{net}\parallel = ma\parallel = (1.8	ext{ kg})(3.0	ext{ m/s}^2) = 5.4	ext{ N}$
Now use Newton's second law and solve for the force of friction:
$F_{net}\parallel = F_{applied} - F_f$
$5.4	ext{ N} = F_{applied} - F_f$
$F_f = F_{applied} - 5.4	ext{ N}$
The only force applied parallel to the incline is gravity, so $F_{applied} = F_g\parallel = 8.82	ext{ N}$
$F_f = 8.82	ext{ N} - 5.4	ext{ N} = 3.42	ext{ N} \approx 3.4	ext{ N}$, so the answer is (C).

Question 7

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. A)5.4 m/s² B)5.2 m/s² C)4.9 m/s² D)5.6 m/s² E)7.9 m/s²

Accepted Answer

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

Question 8

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)2.0 N B)1.4 N C)1.6 N D)2.8 N E)3.4 N

Accepted Answer

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

Question 9

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)14 m
B)11 m
C)19 m
D)24 m
E)29 m

Accepted Answer

The maximum distance the truck can travel without having the box slide can be found using the formula:
$d = \frac{1}{2}at^2$
where $a$ is the acceleration of the truck and $t$ is the time taken. To find $a$, we need to use the coefficient of static friction:
$f_s = \mu_s N$
where $f_s$ is the force of static friction, $\mu_s$ is the coefficient of static friction, and $N$ is the normal force (equal to the weight of the box). Since the box is not moving vertically, the normal force is equal to the weight:
$N = mg$
where $m$ is the mass of the box and $g$ is the acceleration due to gravity. Substituting for $f_s$ and solving for $a$:
$f_s = ma = \mu_s N = \mu_s mg$
$a = \mu_s g = 0.24 	imes 9.8 \approx 2.35 	ext{ m/s}^2$
Substituting into the formula for $d$:
$d = \frac{1}{2}at^2 = \frac{1}{2} (2.35) (3.0)^2 \approx 10.57 \approx 11 	ext{ m}$

Therefore, the best choice is B.

Question 10

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? A)2.8 m/s² B)2.3 m/s² C)1.8 m/s² D)3.3 m/s² E)5.4 m/s²

Accepted Answer

First, we need to resolve the applied force F into its horizontal component: Fcosθ = 20N × cos30° = 17.3N.
Then we need to calculate the force of friction between the block and the surface: Ff = μkN = 0.30 × (3.0kg × 9.81m/s^2) = 8.83N.
The net force acting on the block in the horizontal direction is: Fnet = Fa – Ff = 17.3N – 8.83N = 8.47N.
Finally, we can calculate the acceleration of the block using Newton's second law: Fnet = ma, so a = Fnet/m = 8.47N/3.0kg = 2.8 m/s^2.
Therefore, the correct answer is choice C, 1.8 m/s^2.

Question 11

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)2.6 N
B)1.1 N
C)2.1 N
D)1.6 N
E)3.7 N

Accepted Answer

The answer of In the figure,if F = 2.0 N...

Question 12

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)0.44
B)0.33
C)0.38
D)0.28
E)0.17

Accepted Answer

The answer of The block shown is pulled across the...

Question 13

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)1.8 m/s² B)2.6 m/s² C)1.4 m/s² D)2.2 m/s² E)3.7 m/s²

Accepted Answer

The force acting on the system is F = 22N. 
The larger block experiences a kinetic friction force given by: f1 = μ1* N1 = 0.25 * (Mg - m*g), where N1 = Mg - m*g is the normal force acting on the larger block. 
Similarly, the smaller block experiences a kinetic friction force given by: f2 = μ2 * N2 = 0.4 * (mg), where N2 = mg is the normal force acting on the smaller block. 
The net force acting on the system along the x-direction is given by: 
Fnet = F - f1 - f2 = Ma
where a is the acceleration of the system. 
Substituting the values, we get: 
22 - 0.25*(1*9.8 - 0.2*9.8) - 0.4*(0.2*9.8) = 1*a 
a = 1.4 m/s^2 
Therefore, the magnitude of acceleration of either block is 1.4 m/s^2.

Question 14

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)1.9 N B)2.2 N C)1.3 N D)1.6 N E)3.3 N

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 15

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)14 N
B)12 N
C)16 N
D)18 N
E)25 N

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 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. A)6.0 N B)5.1 N C)5.5 N D)4.6 N E)3.7 N

Accepted Answer

The answer of The three blocks shown are released from...

Question 17

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. 
A)15 N
B)25 N
C)11 N
D)22 N
E)33 N

Accepted Answer

First, we need to find the net force acting on the system. To do this, we can subtract the force of friction from the applied force. 
For the 3.0 kg block: 
F_applied - F_friction = 15 N - (0.30)(3.0 kg)(9.8 m/s^2) = 4.86 N
For the 2.0 kg block: 
F_friction = (0.30)(2.0 kg)(9.8 m/s^2) = 5.88 N
Therefore, the net force acting on the 2.0 kg block is: 
F_net = F_block3 - F_friction = 15 N - 5.88 N = 9.12 N
Thus, the magnitude of the force exerted on the 2.0 kg block by the 3.0 kg block is 9.12 N.

Question 18

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)2.1 m/s² B)2.3 m/s² C)1.9 m/s² D)1.7 m/s² E)3.2 m/s²

Accepted Answer

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

Question 19

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)0.20
B)0.23
C)0.26
D)0.33
E)0.41

Accepted Answer

The answer of A 4.0-kg block slides down a 35°...

Question 20

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?
A)27 N
B)15 N
C)13 N
D)17 N
E)19 N

Accepted Answer

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

Quiz 5: The Laws of Motion

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?

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?

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 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.

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.

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.

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 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 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,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.

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?

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 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 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?

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.

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.

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 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 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?