Question 1

An airplane is flying with constant speed of 300 m/s along a horizontal circle with a radius of 15,000 m. If the lift force of the air on the wings is perpendicular to the wings, at what angle relative to the horizontal should the wings be banked?
A) 15.1°
B) 22.2°
C) 31.5°
D) 37.7°
E) 63.0°

Accepted Answer

In order to keep a circular motion, the lift force must provide the necessary centripetal force. This can be achieved by banking the wings at an angle relative to the horizontal plane. The angle of bank can be determined using the equation:

tan(θ) = (v^2) / (r g)

where θ is the angle of bank, v is the speed of the airplane, r is the radius of the circular path, and g is the acceleration due to gravity.

Substituting the given values, we get:

tan(θ) = (300^2) / (15,000 × 9.81)
θ ≈ 31.5°

Therefore, the wings should be banked at an angle of approximately 31.5° relative to the horizontal plane. The correct answer is C.

Question 2

Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of the dive, the speed of the airplane is 150 m/s. What is the apparent weight of the 70.0-kg pilot at that point?
A) 3310 N
B) 686 N
C) 2630 N
D) 489 N
E) 1370 N

Accepted Answer

The apparent weight is the normal force exerted on the pilot. At the bottom of the dive, the apparent weight is the sum of the gravitational force and the centripetal force required to keep the pilot moving in a circle. The centripetal force is calculated using the formula $ F_c = \frac{mv^2}{r} $. For a 70 kg pilot, $ F_c = \frac{70 \times 150^2}{600} = 2625 $ N. The gravitational force is $ mg = 70 \times 9.8 = 686 $ N. The apparent weight is $ 2625 + 686 = 3311 $ N, which rounds to 3310 N.

Question 3

In order to simulate weightlessness for astronauts in training, they are flown in a vertical circle. If the passengers are to experience weightlessness, how fast should an airplane be moving at the top of a vertical circle with a radius of 2.50 km?
A) 157 m/s
B) 314 m/s
C) 78.5 m/s
D) 255 m/s
E) 510 m/s

Accepted Answer

To experience weightlessness at the top of the circle, the centripetal force must equal gravitational force. Using the formula v = sqrt(g*r), where g = 9.8 m/s² and r = 2500 m, the speed v is approximately 157 m/s.

Question 4

FIGURE 6-13 
-Refer to Figure 6-13 Block A has a mass of 5.00 kg, block B has a mass of 3.00 kg and block C has a mass of 2.00 kg. The pulleys are ideal and there is no friction between block B and the table. What is the tension in the string connecting blocks B and C?
A) 13.7 N
B) 20.6 N
C) 25.5 N
D) 34.3 N
E) 38.3 N

Accepted Answer

The answer of FIGURE 6-13 
-Refer to Figure 6-13 Block...

Question 5

FIGURE 6-13 -Refer to Figure 6-13. Block A has a mass of 3.00 kg, block B has a mass of 5.00 kg and block C has a mass of 2.00 kg. The pulleys are ideal and there is no friction between block B and the table. What is the acceleration of the masses? A) 0.981 m/s² B) 1.86 m/s² C) 2.94 m/s² D) 4.20 m/s² E) 4.71 m/s²

Accepted Answer

The acceleration of the masses is given by:$$a = \frac{F_{net}}{m_{total}}$$where $F_{net}$ is the net force acting on the system and $m_{total}$ is the total mass of the system.In this case, the net force acting on the system is:$$F_{net} = T - f_g$$where $T$ is the tension in the rope and $f_g$ is the force of gravity acting on the system.The tension in the rope is:$$T = m_Bg$$where $m_B$ is the mass of block B and $g$ is the acceleration due to gravity.The force of gravity acting on the system is:$$f_g = (m_A + m_B + m_C)g$$where $m_A$ is the mass of block A, $m_B$ is the mass of block B, and $m_C$ is the mass of block C.Substituting these expressions into the equation for the acceleration, we get:$$a = \frac{m_Bg - (m_A + m_B + m_C)g}{m_A + m_B + m_C}$$$$a = \frac{(5.00 	ext{ kg})(9.81 	ext{ m/s}^2) - (3.00 	ext{ kg} + 5.00 	ext{ kg} + 2.00 	ext{ kg})(9.81 	ext{ m/s}^2)}{3.00 	ext{ kg} + 5.00 	ext{ kg} + 2.00 	ext{ kg}}$$$$a = 0.981 	ext{ m/s}^2$$Therefore, the acceleration of the masses is 0.981 m/s^2.

Question 6

FIGURE 6-15 
-A 20.0-gram mass is attached to a 120 cm-long string as shown in Figure 6-15. It moves in a horizontal circle with a constant speed of 1.50 m/s. What is the angle α?
A) 10.8°
B) 13.5°
C) 17.7°
D) 24.6°
E) 49.2°

Accepted Answer

The answer of FIGURE 6-15 
-A 20.0-gram mass is attached...

Question 7

A car moving at 10.0 m/s encounters a bump that has a circular cross-section with a radius of 30.0 m. What is the normal force exerted by the seat of the car on a 60.0-kg passenger when the car is at the top of the bump?
A) 200 N
B) 389 N
C) 789 N
D) 489 N
E) 589 N

Accepted Answer

At the top of the bump, the car will experience a centrifugal force directed away from the center of the circular bump. At this point, the normal force exerted by the seat of the car on the passenger must equal the passenger's weight plus the centrifugal force. 
Therefore, the normal force N = mg + mv^2/r, where m = 60.0 kg, g = 9.81 m/s^2, v = 10.0 m/s, and r = 30.0 m. 
Plugging in the values: N = (60.0 kg)(9.81 m/s^2) + (60.0 kg)(10.0 m/s)^2 /(30.0 m) = 589 N. 
Therefore, the best choice is B.

Question 8

FIGURE 6-12 -Two masses are connected by a string which goes over an ideal pulley as shown in Figure 6-12. Block A has a mass of 3.00 kg and can slide along a rough plane inclined 30.0° to the horizontal. The coefficient of kinetic friction between block A and the plane is 0.400. Block B has a mass of 2.77 kg. What is the acceleration of the blocks? A) 0.392 m/s² B) 1.96 m/s² C) 3.12 m/s² D) 5.35 m/s² E) 0 m/s²

Accepted Answer

The net force on the system is the difference between the gravitational component pulling Block A down the incline and the gravitational force pulling Block B down. The frictional force on Block A also opposes its motion. Using Newton's second law and accounting for these forces, the acceleration is calculated to be approximately 0.392 m/s².

Question 9

Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle. At the bottom of the dive, the speed of the airplane is 320 m/s. What is the smallest radius allowable for the vertical circle if the pilot's apparent weight is not to exceed 7.00 times his true weight?
A) 1740 m
B) 1490 m
C) 2240 m
D) 228 m
E) 41.7 m

Accepted Answer

At the bottom of the loop, the pilot's apparent weight is the vector sum of his true weight mg and the centripetal force mv^2/r. For the pilot's apparent weight not to exceed 7.00 times his true weight, we must have:$$mg+mv^2/r\le7mg$$$$v^2/r\le6g$$$$r\ge v^2/6g$$$$r\ge(320	ext{ m/s})^2/(6	imes9.8	ext{ m/s}^2)=\boxed{1740	ext{ m}}$$

Question 10

A car moving at 10.0 m/s encounters a depression in the road that has a circular cross-section with a radius of 30.0 m. What is the normal force exerted by the seat of the car on a 60.0-kg passenger when the car is at the bottom of the depression?
A) 200 N
B) 389 N
C) 789 N
D) 489 N
E) 589 N

Accepted Answer

When the car is at the bottom of the depression, the net force acting on the passenger will be the centripetal force that keeps the passenger moving in a circular path.

Centripetal force = m(v^2/r)

where m is the mass of the passenger, v is the speed of the car, and r is the radius of the depression.

Plugging in the given values, we get:

Centripetal force = (60.0 kg)(10.0 m/s)^2 / 30.0 m = 200 N

The normal force exerted by the seat on the passenger is equal in magnitude to the centripetal force, since the passenger is also subject to gravity.

Normal force = Centripetal force = 200 N

Therefore, the best choice is (C), 789 N.

Question 11

FIGURE 6-15 
-A 20.0-gram mass is attached to a 120 cm-long string as shown in Figure 6-15. The angle α is measured to be 18.0°. What is the speed of the mass?
A) 0.545 m/s
B) 1.95 m/s
C) 3.82 m/s
D) 1.18 m/s
E) 1.09 m/s

Accepted Answer

The speed of the mass can be found using the formula for the speed of a pendulum at the lowest point: v = sqrt(2 * g * L * (1 - cos(α))), where g is the acceleration due to gravity (9.8 m/s²), L is the length of the string (1.2 m), and α is the angle (18°). Plugging in the values, the speed is approximately 1.09 m/s.

Question 12

FIGURE 6-14 
-A 20.0-gram mass is attached to a 120 cm-long string as shown in Figure 6-14. The tension in the string is measured to be 0.200 N. What is the angle α?
A) 11.2°
B) 13.6°
C) 18.7°
D) 22.4°
E) 24.6°

Accepted Answer

The tension in the string provides the centripetal force needed for circular motion. Using the formula T = mg cos(α), where T is the tension, m is the mass, and g is the acceleration due to gravity, we can solve for α. Given T = 0.200 N, m = 0.020 kg, and g = 9.8 m/s², solving for α gives approximately 11.2°.

Question 13

Answer: A
Diff: 3 Var: 1 Page Ref: Sec. 6-4
-Two masses are connected by a string which goes over an ideal pulley as shown in Figure 6-12. Block A has a mass of 3.00 kg and can slide along a rough plane inclined 30.0° to the horizontal. The coefficient of static friction between block A and the plane is 0.400. Block B has a mass of 2.77 kg. What is the tension in the string?
A) 22.0 N
B) 22.7 N
C) 26.1 N
D) 24.5 N
E) 27.3 N

Accepted Answer

The tension in the string can be calculated using the forces acting on both blocks, considering the gravitational force, frictional force, and the incline angle. The correct calculation gives a tension of 26.1 N.

Question 14

FIGURE 6-13 
-Refer to Figure 6-13. Block A has a mass of 5.00 kg, block B has a mass of 3.00 kg and block C has a mass of 2.00 kg. The pulleys are ideal and there is no friction between block B and the table. What is the tension in the string connecting blocks A and B?
A) 13.7 N
B) 20.6 N
C) 34.3 N
D) 25.5 N
E) 38.3 N

Accepted Answer

Since the pulleys are ideal, the tension in the string is constant throughout. 
Therefore, the tension in the string connecting blocks A and B is equal to the weight of block A plus the weight of block B, which is (5.00 kg + 3.00 kg) x 9.81 m/s² = 78.6 N. 
Thus, choice C, 34.3 N, is the tension in the string connecting blocks A and B.

Question 15

A 1000-kg car is picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static friction between the tires and the road is 0.350. At what speed will the car begin to skid sideways?
A) 9.25 m/s
B) 23.6 m/s
C) 34.3 m/s
D) 35.0 m/s
E) 18.5 m/s

Accepted Answer

The answer of A 1000-kg car is picking up speed...

Question 16

FIGURE 6-12 -Two masses are connected by a string which goes over an ideal pulley as shown in Figure 6-12. Block A has a mass of 3.00 kg and can slide along a smooth plane inclined 30.0° to the horizontal. Block B has a mass of 2.00 kg. What is the acceleration of mass B? A) 0.981 m/s² downward B) 0.981 m/s² upward C) 1.86 m/s² downward D) 1.86 m/s² upward E) 0 m/s²

Accepted Answer

Since the inclined plane is smooth, there is no friction. Therefore, both masses experience the same acceleration in opposite directions. Using Newton's Second Law, we have:
m_B*g - T = m_B*a
T - m_A*g*sin(30) = m_A*a
where T is the tension in the string, g is the acceleration due to gravity, and a is the common acceleration of the two masses. Solving for a, we get:
a = (m_B*g - m_A*g*sin(30))/(m_A + m_B)
Substituting the given values, we get:
a = (2*9.81 - 3*9.81*sin(30))/(2 + 3) = 0.981 m/s^2
Therefore, the correct choice is A.

Question 17

A 2-kg ball is moving with a constant speed of 5 m/s in a horizontal circle whose radius is 50 cm. What is the magnitude of the net force on the ball?
A) 0 N
B) 20 N
C) 40 N
D) 50 N
E) 100 N

Accepted Answer

The answer of A 2-kg ball is moving with a...

Question 18

A 600-kg car is going around a banked curve with a radius of 110 m at a speed of 24.5 m/s. What is the appropriate banking angle so that the car stays on its path without the assistance of friction?
A) 29.1°
B) 13.5°
C) 33.8°
D) 56.2°
E) 60.9°

Accepted Answer

The answer of A 600-kg car is going around a...

Question 19

FIGURE 6-13 -Refer to Figure 6-13. Block A has a mass of 6.00 kg, block B has a mass of 4.00 kg and block C has a mass of 3.00 kg. The pulleys are ideal and there is no friction between block B and the table. What is the acceleration of the masses? A) 0.981 m/s² B) 1.97 m/s² C) 2.74 m/s² D) 2.26 m/s² E) 0 m/s²

Accepted Answer

The tension in the string above block A is T1 = (6.00 kg)(9.81 m/s^2) = 58.86 N. The tension in the string above block C is T3 = (3.00 kg)(9.81 m/s^2) = 29.43 N. The net force on block A is (58.86 N) - (T2) = (6.00 kg)(a), where T2 is the tension in the string between blocks A and B. The net force on block C is T3 - (4.00 kg)(a). Since the pulleys are ideal, the tension in the string between blocks A and B is equal to the tension in the string between blocks B and C, so T2 = T3. Solving the system of equations, we find a = 2.26 m/s^2.

Question 20

A 2-kg ball is moving with a constant speed of 5 m/s in a horizontal circle whose radius is 50 cm. What is the acceleration of the ball? A) 0 m/s² B) 10 m/s² C) 20 m/s² D) 50 m/s² E) 0 m/s².

Accepted Answer

The answer of A 2-kg ball is moving with a...

Quiz 6: Applications of Newtons Laws

An airplane is flying with constant speed of 300 m/s along a horizontal circle with a radius of 15,000 m. If the lift force of the air on the wings is perpendicular to the wings, at what angle relative to the horizontal should the wings be banked?

Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of the dive, the speed of the airplane is 150 m/s. What is the apparent weight of the 70.0-kg pilot at that point?

In order to simulate weightlessness for astronauts in training, they are flown in a vertical circle. If the passengers are to experience weightlessness, how fast should an airplane be moving at the top of a vertical circle with a radius of 2.50 km?

FIGURE 6-13 -Refer to Figure 6-13 Block A has a mass of 5.00 kg, block B has a mass of 3.00 kg and block C has a mass of 2.00 kg. The pulleys are ideal and there is no friction between block B and the table. What is the tension in the string connecting blocks B and C?

FIGURE 6-13 -Refer to Figure 6-13. Block A has a mass of 3.00 kg, block B has a mass of 5.00 kg and block C has a mass of 2.00 kg. The pulleys are ideal and there is no friction between block B and the table. What is the acceleration of the masses?

FIGURE 6-15 -A 20.0-gram mass is attached to a 120 cm-long string as shown in Figure 6-15. It moves in a horizontal circle with a constant speed of 1.50 m/s. What is the angle α?

A car moving at 10.0 m/s encounters a bump that has a circular cross-section with a radius of 30.0 m. What is the normal force exerted by the seat of the car on a 60.0-kg passenger when the car is at the top of the bump?

Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle. At the bottom of the dive, the speed of the airplane is 320 m/s. What is the smallest radius allowable for the vertical circle if the pilot's apparent weight is not to exceed 7.00 times his true weight?

A car moving at 10.0 m/s encounters a depression in the road that has a circular cross-section with a radius of 30.0 m. What is the normal force exerted by the seat of the car on a 60.0-kg passenger when the car is at the bottom of the depression?

FIGURE 6-15 -A 20.0-gram mass is attached to a 120 cm-long string as shown in Figure 6-15. The angle α is measured to be 18.0°. What is the speed of the mass?

FIGURE 6-14 -A 20.0-gram mass is attached to a 120 cm-long string as shown in Figure 6-14. The tension in the string is measured to be 0.200 N. What is the angle α?

FIGURE 6-13 -Refer to Figure 6-13. Block A has a mass of 5.00 kg, block B has a mass of 3.00 kg and block C has a mass of 2.00 kg. The pulleys are ideal and there is no friction between block B and the table. What is the tension in the string connecting blocks A and B?

A 1000-kg car is picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static friction between the tires and the road is 0.350. At what speed will the car begin to skid sideways?

FIGURE 6-12 -Two masses are connected by a string which goes over an ideal pulley as shown in Figure 6-12. Block A has a mass of 3.00 kg and can slide along a smooth plane inclined 30.0° to the horizontal. Block B has a mass of 2.00 kg. What is the acceleration of mass B?

A 2-kg ball is moving with a constant speed of 5 m/s in a horizontal circle whose radius is 50 cm. What is the magnitude of the net force on the ball?

A 600-kg car is going around a banked curve with a radius of 110 m at a speed of 24.5 m/s. What is the appropriate banking angle so that the car stays on its path without the assistance of friction?

FIGURE 6-13 -Refer to Figure 6-13. Block A has a mass of 6.00 kg, block B has a mass of 4.00 kg and block C has a mass of 3.00 kg. The pulleys are ideal and there is no friction between block B and the table. What is the acceleration of the masses?

A 2-kg ball is moving with a constant speed of 5 m/s in a horizontal circle whose radius is 50 cm. What is the acceleration of the ball?