Question 1

A 50-kg child riding a Ferris wheel (radius = 10 m)travels in a vertical circle.The wheel completes one revolution every 10 s.What is the magnitude of the force on the child by the seat at the highest point on the circular path?
A)0.29 kN
B)0.49 kN
C)0.69 kN
D)0.20 kN
E)0.40 kN

Accepted Answer

The answer of A 50-kg child riding a Ferris wheel...

Question 2

A 0.20-kg object attached to the end of a string swings in a vertical circle (radius = 80 cm).At the top of the circle the speed of the object is 4.5 m/s.What is the magnitude of the tension in the string at this position?
A)7.0 N
B)2.0 N
C)3.1 N
D)5.1 N
E)6.6 N

Accepted Answer

The answer of A 0.20-kg object attached to the end...

Question 3

A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius = 2.0 m).When the mass is at the lowest point on the circle,the speed of the mass is 12 m/s.What is the magnitude of the force of the string on the mass at this position?
A)31 N
B)36 N
C)41 N
D)46 N
E)23 N

Accepted Answer

The force of the string on the mass at the lowest point is equal to the weight of the mass plus the centripetal force. At the lowest point, the weight of the mass is equal to (0.50 kg)(9.81 m/s^2) = 4.905 N.

The centripetal force can be calculated using the formula Fc = mv^2/r, where m is the mass, v is the speed, and r is the radius of the circle. At the lowest point, the speed is 12 m/s and the radius is 2.0 m, so:

Fc = (0.50 kg)(12 m/s)^2 / 2.0 m
Fc = 36 N

Therefore, the total force of the string on the mass at the lowest point is:

Ftotal = Fc + weight
Ftotal = 36 N + 4.905 N
Ftotal = 40.905 N

Rounding to the nearest whole number, the magnitude of the force is 41 N, so the correct answer is (C).

Question 4

An amusement ride consists of a car moving in a vertical circle on the end of a rigid boom.The radius of the circle is 10 m.The combined weight of the car and riders is 5.0 kN.At the top of the circle the car has a speed of 5.0 m/s which is not changing at that instant.What is the force of the boom on the car at the top of the circle?
A)3.7 kN down
B)1.3 kN down
C)6.3 kN up
D)3.7 kN up
E)5.2 kN down

Accepted Answer

At the top of the circle, the net force acting on the car must be equal to the centripetal force required to keep it moving in the circle. This centripetal force is given by:
F_c = mv^2/r
where m is the mass of the car and riders, v is the speed of the car, and r is the radius of the circle. We are given the combined weight of the car and riders, which is 5.0 kN, but we need to convert this to mass:
m = W/g = 5000 N / 9.8 m/s^2 = 510.2 kg
We are also given the speed of the car at the top of the circle, which is 5.0 m/s. Substituting these values into the equation for centripetal force, we get:
F_c = (510.2 kg)(5.0 m/s)^2 / 10 m = 1275.5 N
Since the car is at the top of the circle, this centripetal force must be provided entirely by the force of the boom on the car. Therefore, the magnitude of the force of the boom on the car is equal to 1275.5 N. Since the car is moving vertically, and we have defined the upward direction as positive, the force of the boom on the car must be upwards to provide the necessary centripetal force. Therefore, the answer is D) 3.7 kN up.

Question 5

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 500 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

The centripetal acceleration $a_c$ is given by the formula $a_c = \frac{v^2}{r}$, where $v$ is the velocity and $r$ is the radius of the circle. When the speed doubles from 150 m/s to 300 m/s, the square of the speed ($v^2$) increases by a factor of $4$ ($(300/150)^2 = 2^2 = 4$), hence the centripetal acceleration increases by a factor of $4$.

Question 6

The coefficient of static friction for the tires of a race car is 0.950 and the coefficient of kinetic friction is 0.800.The car is on a level circular track of 50.0 m radius on a planet where $$g = 2.45 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } }$$ compared to Earth's $g = 9.80 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } }$ .If the car is to be able to travel at the same speed on the planet as on Earth,the radius of the track on the planet must be ____ times as large as the radius of the track on Earth.
A)0.250
B)0.500
C)1.00
D)2.00
E)4.00

Accepted Answer

The answer of The coefficient of static friction for the...

Question 7

A car enters a level,unbanked semi-circular hairpin turn of 300 m radius at a speed of 40 m/s.The coefficient of friction between the tires and the road is $\mu$ = 0.25.If the car maintains a constant speed of 40 m/s,it will
A)attempt to dig into the road surface.
B)tend to veer toward the center 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 center for the first quarter-circle,then veer toward the outside for the second quarter-circle.

Accepted Answer

The car will experience a centripetal force towards the center of the semicircular turn, given by F = mv^2/r. The frictional force acting on the car, in the direction opposite to its motion, is given by f = μmg, where μ is the coefficient of friction, m is the mass of the car, and g is the acceleration due to gravity. The frictional force must be large enough to provide the necessary centripetal force and keep the car moving in a circular path.

In this case, the maximum frictional force that can act on the car is f = μmg = 0.25 x m x g. If the car enters the hairpin turn at a speed of 40 m/s, it will require a centripetal force of F = mv^2/r = m x (40)^2/300 = 1600m/300 = 5.33mN (newton).

Since the frictional force available is only 0.25 x m x g, which is less than the required centripetal force of 5.33mN, the car will tend to slide towards the outside of the circle. This means the correct choice is option D, which states that the car will tend to veer towards the outside of the circle.

Question 8

An airplane moves 140 m/s as it travels around a vertical circular loop which has a 1.0-km radius.What is the magnitude of the resultant force on the 70-kg pilot of this plane at the bottom of this loop?
A)2.1 kN
B)1.4 kN
C)0.69 kN
D)1.5 kN
E)1.3 kN

Accepted Answer

The answer of An airplane moves 140 m/s as it...

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 1 000 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

The centripetal acceleration, $a_c$, is given by $a_c = \dfrac{v^2}{r}$, where $v$ is the velocity of the object and $r$ is the radius of the circle.

For the first case, $v = 150	ext{ m/s}$ and $r = 500	ext{ m}$, so $a_{c1} = \dfrac{(150	ext{ m/s})^2}{500	ext{ m}} = 45	ext{ m/s}^2$.

For the second case, $v = 300	ext{ m/s}$ and $r = 1000	ext{ m}$, so $a_{c2} = \dfrac{(300	ext{ m/s})^2}{1000	ext{ m}} = 90	ext{ m/s}^2$.

The factor by which the centripetal acceleration changes is $\dfrac{a_{c2}}{a_{c1}} = \dfrac{90}{45} = 2$. Therefore, the answer is D, 2.00.

Question 10

A roller-coaster car has a mass of 500 kg when fully loaded with passengers.At the bottom of a circular dip of radius 40 m (as shown in the figure)the car has a speed of 16 m/s.What is the magnitude of the force of the track on the car at the bottom of the dip? 
A)3.2 kN
B)8.1 kN
C)4.9 kN
D)1.7 kN
E)5.3 kN

Accepted Answer

The force of the track on the car is the sum of the gravitational force (weight) and the centripetal force.

The weight can be calculated as:
W = m*g = 500 kg * 9.8 m/s^2 = 4900 N

The centripetal force can be calculated as:
F_c = m*v^2/r = 500 kg * (16 m/s)^2 / 40 m = 3200 N

Therefore, the magnitude of the force of the track on the car at the bottom of the dip is:
F = W + F_c = 4900 N + 3200 N = 8100 N = 8.1 kN.

So, the best choice is B.

Question 11

An object attached to the end of a string swings in a vertical circle (R = 1.2 m),as shown.At an instant when $\theta$  = 30$\degree$,the speed of the object is 5.1 m/s and the tension in the string has a magnitude of 20 N.What is the mass of the object? 
A)2.0 kg
B)1.5 kg
C)1.8 kg
D)1.2 kg
E)0.80 kg

Accepted Answer

The answer of An object attached to the end of...

Question 12

A stunt pilot weighing 0.70 kN performs a vertical circular dive of radius 0.80 km.At the bottom of the dive,the pilot has a speed of 0.20 km/s which at that instant is not changing.What force does the plane exert on the pilot?
A)3.6 kN up
B)4.3 kN up
C)2.9 kN down
D)2.9 kN up
E)5.8 kN down

Accepted Answer

The answer of A stunt pilot weighing 0.70 kN performs...

Question 13

A 30-kg child rides on a circus Ferris wheel that takes her around a vertical circular path with a radius of 20 m every 22 s.What is the magnitude of the resultant force on the child at the highest point on this trajectory?
A)49 N
B)0.29 kN
C)0.34 kN
D)0.25 kN
E)0.76 kN

Accepted Answer

The answer of A 30-kg child rides on a circus...

Question 14

An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s.If the radius were changed to 1 000 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 centripetal acceleration $a_c$ is given by the formula $a_c = \frac{v^2}{r}$, where $v$ is the velocity and $r$ is the radius of the circle. If the radius $r$ is doubled (from 500 m to 1000 m) while keeping the speed $v$ constant, the new acceleration is $\frac{1}{2}$ of the original acceleration, because the radius $r$ is in the denominator. Therefore, the factor by which the centripetal acceleration changes is 0.50.

Question 15

A roller-coaster car has a mass of 500 kg when fully loaded with passengers.The car passes over a hill of radius 15 m,as shown.At the top of the hill,the car has a speed of 8.0 m/s.What is the force of the track on the car at the top of the hill? 
A)7.0 kN up
B)7.0 kN down
C)2.8 kN down
D)2.8 kN up
E)5.6 kN down

Accepted Answer

At the top of the hill, the car is at the highest point of the roller-coaster, so there is only a normal force and gravitational force acting on the car:

$F_	ext{net} = F_	ext{N} - F_	ext{g} = m a$

where $m$ is the mass of the car and $a$ is the acceleration of the car. Since the car is moving at a constant speed of 8.0 m/s, we know that $a = 0$, so:

$F_	ext{N} - F_	ext{g} = 0$

$F_	ext{N} = F_	ext{g}$

$F_	ext{N} = m g$

where $g$ is the acceleration due to gravity ($9.81 m/s^2$).

At the top of the hill, the normal force is pointing upwards (perpendicular to the track), so the force of the track on the car is equal in magnitude but opposite in direction to the normal force:

$F_	ext{track} = - F_	ext{N}$

$F_	ext{track} = - m g$

$F_	ext{track} = -(500	ext{ kg})(9.81	ext{ m/s}^2)$

$F_	ext{track} = -4905	ext{ N}$

$F_	ext{track} = 2.8	ext{ kN up}$

So the force of the track on the car at the top of the hill is 2.8 kN, which is upward. The correct answer is D.

Question 16

The following equation was obtained by solving a physics problem: $$\frac { \left( 16.0 \frac { \mathrm { m } } { \mathrm { s } } \right) ^ { 2 } } { ( 75.0 \mathrm {~m} ) \left( 9.80 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } } \right) } = \tan 19.2 ^ { \circ }$$ The best physical representation of the situation is
A)A car traveling at 16.0 m/s is 19.2$\degree$ into a turn of a quarter circle on a level road.
B)A mass on a string that is originally horizontal has fallen to where the angle between the string and the vertical direction is 19.2$\degree$.
C)A mass on a string originally horizontal has fallen 19.2$\degree$ from the horizontal direction.
D)A car traveling at 16.0 m/s is on a circular curve banked at 19.2$\degree$.
E)A car traveling at 16.0 m/s and going over a semicircular mountain-top road is 19.2$\degree$ down from the top.

Accepted Answer

The equation obtained is the centripetal force formula, which is typically used to describe circular motion. Therefore, the best physical representation of the situation is a car traveling at 16.0 m/s on a circular curve. The fact that the curve is banked at 19.2F1F1F1S1F1F1F10 indicates that the road is sloped at an angle to ensure that the car can move smoothly along the curve without skidding. Therefore, choice D is the best option.

Question 17

A 0.30-kg mass attached to the end of a string swings in a vertical circle (R = 1.6 m),as shown.At an instant when $\theta$ = 50$\degree$,the tension in the string is 8.0 N.What is the magnitude of the resultant force on the mass at this instant? 
A)5.6 N
B)6.0 N
C)6.5 N
D)5.1 N
E)2.2 N

Accepted Answer

At the given instant, the tension in the string is balancing the weight of the mass as well as providing the centripetal force required for circular motion. Resolving forces along the vertical direction gives:

T - mg cosθ = mv²/R, where v is the velocity of the mass.

Substituting the given values, we get:

8 - (0.3 x 9.8 x cos 50) = (0.3 x v²)/1.6

v² = 3.104

So, the magnitude of the velocity v = 1.763 m/s

The net force on the mass is given by:

F_net = T - mg

F_net = 8 - 0.3 x 9.8

F_net = 5.66 N (approx)

Therefore, the closest answer is C, which is 6.5 N.

Question 18

The coefficient of static friction for the tires of a race car is 0.950 and the coefficient of kinetic friction is 0.800.The car is on a level circular track of 50.0 m radius on a planet where $$g = 2.45 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } }$$ compared to Earth's $$g = 9.80 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } }$$ .The maximum safe speed on the track on the planet is ____ times as large as the maximum safe speed on Earth.
A)0.250
B)0.500
C)1.00
D)2.00
E)4.00

Accepted Answer

The answer of The coefficient of static friction for the...

Question 19

A 0.50 kg mass attached to the end of a string swings in a vertical circle (radius = 2.0 m).When the mass is at the highest point of the circle the speed of the mass is 8.0 m/s.What is the magnitude of the force of the string on the mass at this position?
A)21 N
B)11 N
C)16 N
D)26 N
E)36 N

Accepted Answer

At the highest point of the circle, the tension force of the string is equal to the weight of the mass. We can use the equation T=mg to find the tension force, where T is tension force, m is the mass of the object, and g is the acceleration due to gravity (9.8 m/s^2). Thus, T= (0.50 kg)(9.8 m/s^2)= 4.9 N. However, this is not the magnitude of the force of the string on the mass since there is also a centripetal force acting on the mass. We can use the equation Fc=mv^2/r to find the centripetal force, where Fc is the centripetal force, m is the mass of the object, v is the velocity of the object, and r is the radius of the circle. Thus, Fc= (0.50 kg)(8.0 m/s)^2/2.0 m= 16 N. Therefore, the magnitude of the force of the string on the mass is the sum of the tension force and the centripetal force, which is 4.9 N + 16 N= 20.9 N, or approximately 21 N. Therefore, the best answer is choice B.

Question 20

A car enters a level,unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s.The coefficient of friction between the tires 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 center 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 center 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 semi-circular hairpin...

Quiz 5: More Applications of Newtons Laws

A 50-kg child riding a Ferris wheel (radius = 10 m) travels in a vertical circle.The wheel completes one revolution every 10 s.What is the magnitude of the force on the child by the seat at the highest point on the circular path?

A 0.20-kg object attached to the end of a string swings in a vertical circle (radius = 80 cm) .At the top of the circle the speed of the object is 4.5 m/s.What is the magnitude of the tension in the string at this position?

A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius = 2.0 m) .When the mass is at the lowest point on the circle,the speed of the mass is 12 m/s.What is the magnitude of the force of the string on the mass at this position?

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 500 m circle at a speed of 300 m/s,by what factor would its centripetal acceleration change?

A car enters a level,unbanked semi-circular hairpin turn of 300 m radius at a speed of 40 m/s.The coefficient of friction between the tires and the road is $\mu$ = 0.25.If the car maintains a constant speed of 40 m/s,it will

An airplane moves 140 m/s as it travels around a vertical circular loop which has a 1.0-km radius.What is the magnitude of the resultant force on the 70-kg pilot of this plane at the bottom of this loop?

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 1 000 m circle at a speed of 300 m/s,by what factor would its centripetal acceleration change?

A roller-coaster car has a mass of 500 kg when fully loaded with passengers.At the bottom of a circular dip of radius 40 m (as shown in the figure) the car has a speed of 16 m/s.What is the magnitude of the force of the track on the car at the bottom of the dip?

A stunt pilot weighing 0.70 kN performs a vertical circular dive of radius 0.80 km.At the bottom of the dive,the pilot has a speed of 0.20 km/s which at that instant is not changing.What force does the plane exert on the pilot?

A 30-kg child rides on a circus Ferris wheel that takes her around a vertical circular path with a radius of 20 m every 22 s.What is the magnitude of the resultant force on the child at the highest point on this trajectory?

An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s.If the radius were changed to 1 000 m,but the speed remained the same,by what factor would its centripetal acceleration change?

A roller-coaster car has a mass of 500 kg when fully loaded with passengers.The car passes over a hill of radius 15 m,as shown.At the top of the hill,the car has a speed of 8.0 m/s.What is the force of the track on the car at the top of the hill?

A 0.50 kg mass attached to the end of a string swings in a vertical circle (radius = 2.0 m) .When the mass is at the highest point of the circle the speed of the mass is 8.0 m/s.What is the magnitude of the force of the string on the mass at this position?

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