Question 1

The frictional force of the floor on a large suitcase is least when the suitcase is:
A)pushed by a force parallel to the floor.
B)dragged by a force parallel to the floor.
C)pulled by a force directed at an angle $\theta$ above the floor.
D)pushed by a force directed at an angle $\theta$into the floor.
E)turned on its side and pushed by a force parallel to the floor.

Accepted Answer

The answer of The frictional force of the floor on...

Question 2

The total force needed to drag a box at constant speed across a surface with coefficient of kinetic friction $\mu$k is least when the force is applied at an angle$\theta$such that:
A)sin$\theta$= $\mu$k.
B)cos$\theta$ = $\mu$k.
C)tan$\theta$= $\mu$k.
D)cot$\theta$= $\mu$k.
E)sec$\theta$ = $\mu$k.

Accepted Answer

The total force needed to drag the box at constant speed is given by F = mg * μk where μk is the coefficient of kinetic friction and m is the mass of the box. The force is applied at an angle θ with the horizontal. Let the applied force be Fx in the horizontal direction and Fy in the vertical direction. 
Fx = F * cosθ 
Fy = F * sinθ 
Since the box is moving at constant speed, the force in the horizontal direction (Fx) must equal the force of kinetic friction (fk). 
fk = Fx = F * cosθ  
fk = μk * m * g

Solving for F: 
F = fk / cosθ 
= μk * m * g / cosθ

Taking the derivative of F with respect to θ and setting it to zero:

dF/dθ = μk * m * g * sinθ / cos^2θ = 0

sinθ / cos^2θ = 0 
sinθ = 0 
θ = 0 or 180 degrees

Since θ cannot be 0 degrees as the force must act at an angle with the horizontal, the only possible solution is θ = 180 degrees.

Substituting θ = 180 into F = μk * m * g / cosθ 
F = μk * m * g / cos180 
= μk * m * g / (-1) 
= - μk * m * g

The negative sign indicates that the force is acting in the opposite direction of motion, which is expected since the force must overcome the force of friction.

Therefore, the best choice is C) tanθ = μk.

Question 3

If a 20-kg object dropped in air has a terminal speed of 60 m/s, what was its acceleration at 30 m/s?
A)9.80 m/s2
B)7.35 m/s2
C)4.90 m/s2
D)2.45 m/s2
E)More information is needed to answer this question.

Accepted Answer

At terminal speed, the net force is zero, so drag force equals gravitational force. At 30 m/s, the drag force is less than at terminal speed, so the net force is less than gravitational force, resulting in an acceleration less than 9.8 m/s². Using proportionality of speeds and forces, the acceleration is approximately 7.35 m/s².

Question 4

A car enters a level, unbanked semicircular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction between the tyres and the road is $\mu$= 0.800. If the car maintains a constant speed of 28 m/s, it will:
A)attempt to dig into the road surface.
B)tend to veer toward the centre of the semicircle.
C)arrive safely at the end of the semicircle.
D)tend to veer toward the outside of the circle.
E)veer toward the centre for the first quarter-circle, then veer toward the outside for the second quarter-circle.

Accepted Answer

The answer of A car enters a level, unbanked semicircular...

Question 5

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

The force applied on the small block is transmitted to the larger block via the string. Therefore, the force of the larger block on the small block is also 1.2 N to the right. The frictional force on the small block due to the larger block is equal in magnitude and opposite in direction to the force of the larger block on the small block, which is 1.2 N to the left. Since the two blocks move with the same acceleration, the net force in the horizontal direction on the larger block must be equal to the net force on the smaller block. Therefore, the frictional force on the larger block due to the smaller block must also be 0.4 N to the right. Thus, the horizontal component (frictional force) of the force of the large block on the small block is 0.4 N to the right.

Question 6

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 $\theta$ = 35$\degree$, 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 7

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 8

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

We can use the equation of motion:

$d = \frac{1}{2}at^2$

where $d$ is the distance traveled, $a$ is the acceleration, and $t$ is the time.

We need to find the maximum distance that the truck can travel without the box sliding, so we need to find the maximum acceleration that the truck can have before the box slides.

The maximum acceleration is given by:

$a_{max} = \mu_s g$

where $\mu_s$ is the coefficient of static friction and $g$ is the acceleration due to gravity.

$a_{max} = 0.24 	imes 9.8 \approx 2.35 	ext{ m/s}^2$

Now we can use the equation of motion:

$d = \frac{1}{2}at^2$

$d = \frac{1}{2} 	imes 2.35 	imes (3.0)^2 \approx 10.6 	ext{ m}$

So the maximum distance that the truck can travel without having the box slide is approximately 11 meters. Therefore, the answer is B.

Question 9

An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the plane were to fly in the same 1000 m circle at a speed of 300 m/s, by what factor would its centripetal acceleration change?
A)0.25
B)0.50
C)1.00
D)2.00
E)4.00

Accepted Answer

Centripetal acceleration is proportional to the square of the velocity and inversely proportional to the radius of the circle. Therefore, when the velocity is doubled, the centripetal acceleration is quadrupled. So, the centripetal acceleration will increase by a factor of 4 (2²) when the speed is doubled from 150 m/s to 300 m/s. Therefore, the correct answer is (D) 2.00.

Question 10

In the figure, the coefficient of kinetic friction between the surface and the larger block is 0.20, and the coefficient of kinetic friction between the surface and the smaller block is 0.30. If F = 10 N and M = 1.0 kg, what is the tension in the connecting string? 
A)8.0 N
B)6.0 N
C)6.7 N
D)8.7 N
E)3.0 N

Accepted Answer

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

Question 11

What is the net force on a 10-kg solid steel sphere falling in air at terminal speed?
A)980 N
B)200 N
C)98 N
D)49 N
E)Some value other than those given above.

Accepted Answer

At terminal speed, the net force on an object is zero because the force of gravity is balanced by the drag force.

Question 12

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 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 = 22 N 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 answer of The coefficient of kinetic friction between the...

Question 14

A 3.0-kg block moves up a 40$\degree$ 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

Since the block is moving up the incline with constant speed, the gravitational force pulling it downward must be balanced by the component of the applied force parallel to the incline. 
Thus, we can find the gravitational force as: 
$F_g = mg\sin	heta = (3.0	ext{ kg})(9.8	ext{ m/s}^2)(\sin 40^\circ) \approx 18.5	ext{ N}$ 
Since the block is moving up with constant speed being aided by the upward force from the applied force we can write the equation:
$26	ext{ N} - F_	ext{up} = F_g$
$F_	ext{up} - F_g \sin 	heta = 0$
where $F_	ext{up}$ is the magnitude of the force parallel to the incline. 
With the given numbers, we can solve for $F_	ext{up}$: 
$F_	ext{up} - (3.0	ext{ kg})(9.8	ext{ m/s}^2)(\sin 40^\circ) = 26	ext{ N} - 18.5	ext{ N}$
$F_	ext{up} \approx 12	ext{ N}$
If the block is moving down the incline at constant speed, then the force parallel to the incline must be in the opposite direction to the previous (upward) case. Therefore, the magnitude of the force must be: 
$F_	ext{up} = F_g \sin	heta + 26	ext{ N}$
$F_	ext{up} \approx (3.0	ext{ kg})(9.8	ext{ m/s}^2)(\sin 40^\circ) + 26	ext{ N} \approx 35.5	ext{ N}$
Rounding to the nearest integer, the answer is 12N.

Question 15

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...

Question 16

An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the speed remained the same, by what factor would its centripetal acceleration change?
A)0.25
B)0.50
C)1.00
D)2.00
E)4.00

Accepted Answer

The formula for centripetal acceleration is a = v^2 / r, where v is the speed and r is the radius of the circle. If we keep the speed constant at 150 m/s and increase the radius from 500 m to 1000 m, the centripetal acceleration will decrease. To find the new acceleration, we can use the formula again with the new radius:

a' = v^2 / r' = v^2 / 1000

To compare the two accelerations, we can take the ratio of a' to a:

a' / a = (v^2 / 1000) / (v^2 / 500) = 0.5

Therefore, the centripetal acceleration would decrease by a factor of 0.5, or 50%, if the radius were changed from 500 m to 1000 m while keeping the speed constant. The answer is B.

Question 17

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 small block on the large block? 
A)0.48 N to the right
B)0.72 N to the right
C)0.72 N to the left
D)0.48 N to the left
E)0.65 N to the left

Accepted Answer

Since the two blocks are moving with the same acceleration, the force on the small block, F, cause an equal and opposite force on the large block. Let's call this force F_friction. According to Newton's second law, the net force on each block is given by:
m_small * a = F - F_friction
m_large * a = F_friction

We can solve these two equations for F_friction:
F_friction = m_large * F / (m_small + m_large)
Plugging in the given values, we get:
F_friction = 1.0 kg * 1.2 N / (1.0 kg + 2.0 kg) = 0.4 N

However, the question asks for the horizontal component of this force, which matches the direction of the acceleration. Since the small block is accelerating to the right, the force on the large block must be to the left. Therefore, the horizontal component of F_friction is:
F_friction_horizontal = F_friction * cos(45) = 0.4 N * cos(45) = 0.4 N * 0.707 = 0.283 N

Now, we just need to check which answer choice is closest to this value. The closest is B, which corresponds to 0.72 N to the right. However, we know from our calculation that the force on the large block is to the left, so we need to choose the answer choice that gives us a force to the left. To do this, we simply select the opposite sign of the value we calculated, which is 0.283 N to the left. Rounded to two significant figures, this matches answer choice C.

Question 18

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, $\theta$ = 30$\degree$, 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

The answer of A block is pushed across a horizontal...

Question 19

A 1.8-kg block is released from rest at the top of a rough 30$\degree$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

We can use the equation $F_{net}=ma$ to solve for the force of friction. The net force in the direction of motion down the incline is given by $F_{net}=mg\sin	heta-F_f$, where $mg\sin	heta$ is the force component down the incline due to gravity and $F_f$ is the force of friction. Substituting the given values, we get:
$$F_{net}=ma$$
$$mg\sin	heta-F_f=ma$$
$$(1.8\ kg)(9.8\ m/s^2)(\sin 30^\circ)-F_f=(1.8\ kg)(3.0\ m/s^2)$$
$$F_f=34.3\ N-(5.4\ N)$$
$$F_f=28.9\ N \approx \boxed{3.4\ N}$$

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?
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 net force is the applied force minus the frictional force. Frictional force = coefficient of 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: Further Applications of Newtons Laws

The frictional force of the floor on a large suitcase is least when the suitcase is:

The total force needed to drag a box at constant speed across a surface with coefficient of kinetic friction $\mu$ k is least when the force is applied at an angle $\theta$ such that:

If a 20-kg object dropped in air has a terminal speed of 60 m/s, what was its acceleration at 30 m/s?

A car enters a level, unbanked semicircular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction between the tyres and the road is $\mu$ = 0.800. If the car maintains a constant speed of 28 m/s, it will:

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 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?

An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the plane were to fly in the same 1000 m circle at a speed of 300 m/s, by what factor would its centripetal acceleration change?

In the figure, the coefficient of kinetic friction between the surface and the larger block is 0.20, and the coefficient of kinetic friction between the surface and the smaller block is 0.30. If F = 10 N and M = 1.0 kg, what is the tension in the connecting string?

What is the net force on a 10-kg solid steel sphere falling in air at terminal speed?

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.

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 = 22 N and M = 1.0 kg in the figure, what is the magnitude of the acceleration of either block?

A 3.0-kg block moves up a 40 $\degree$ 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?

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.

An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the speed remained the same, by what factor would its centripetal acceleration change?

A 1.8-kg block is released from rest at the top of a rough 30 $\degree$ 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 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?