Question

practice: a car drives around a curve with a radius of 50 meters at a velocity of 55 mph. a) imbalance? yes or no
an 80 - kg skydiver is falling downward at a terminal velocity of 125 mph. a) imbalance? yes or no b) what is the magnitude of the drag force from the air on the skydiver?
3 - 2) an 80 - kg skydiver is falling downward at a speed of 40 m/sec, speeding up at a rate of 2 m/sec every second. a) imbalance? yes or no b) what is the magnitude of the drag force from the air on the skydiver?
3 - 6) a person is standing on a scale in an elevator that is moving downward at 4 m/sec and slowing down at a rate of 1 m/sec each second. the person weighs 800 newtons. a) imbalance? yes or no b) in units of newtons, what does the scale read?
3 - 11) a 46 - gram golf ball is shot straight up with a slingshot. one second after it is released, the ball is traveling straight upward at a speed 30 m/sec and the force due to the air drag is 0.14 newtons. a) imbalance? yes or no b) what is the acceleration of the ball? give the magnitude and direction.
3 - 13) a 70 - kg physics student is standing on a scale in an elevator. the scale reads 660 newtons. a) imbalance? yes or no b) what is the acceleration of the elevator? (magnitude and direction). c) can the elevator be moving upward? why or why not?
3 - 17) a 1400 - kg car is parked on a hill that is tilted 8 degrees up from the horizontal. a) imbalance? yes or no b) what is the magnitude of the frictional force on the tires from the street surface? c) what is the minimum coefficient of static friction between the street surface and the tires that will prevent the car from sliding down the hill?

Explanation:

Step1: A car moving in a circular - path needs a centripetal force.

Since the car is moving in a curved path, there is a net force acting towards the center of the curve. So, there is an imbalance.
- # Answer:
Yes
2. **For the skydiver at terminal velocity (80 - kg, 125 mph)**:
- **a) Imbalance?**
- # Explanation:
## Step1: At terminal velocity, the net force on the skydiver is zero.
The gravitational force is balanced by the drag force. So, there is no imbalance.
- # Answer:
No
- **b) Magnitude of the drag force**:
- # Explanation:
## Step1: At terminal velocity, \(F_{drag}=mg\).
Given \(m = 80\ kg\) and \(g = 9.8\ m/s^{2}\), then \(F_{drag}=80\times9.8\).
\(F_{drag}=784\ N\)
- # Answer:
784 N
3. **For the 80 - kg skydiver falling at 40 m/s and accelerating at 2 m/s²**:
- **a) Imbalance?**
- # Explanation:
## Step1: Since there is an acceleration, there is a net force.
According to Newton's second law \(F = ma\), if \(a eq0\), there is an imbalance.
- # Answer:
Yes
- **b) Magnitude of the drag force**:
- # Explanation:
## Step1: Apply Newton's second law \(F_{net}=mg - F_{drag}=ma\).
Rearrange to find \(F_{drag}=mg - ma\).
Given \(m = 80\ kg\), \(g = 9.8\ m/s^{2}\), and \(a = 2\ m/s^{2}\).
\(F_{drag}=80\times9.8-80\times2\).
\(F_{drag}=80\times(9.8 - 2)=80\times7.8 = 624\ N\)
- # Answer:
624 N
4. **For the person in the elevator (weighs 800 N, moving down at 4 m/s and slowing down at 1 m/s²)**:
- **a) Imbalance?**
- # Explanation:
## Step1: Since the elevator is decelerating, there is a net force.
Deceleration implies a non - zero acceleration, so there is an imbalance.
- # Answer:
Yes
- **b) Scale reading**:
- # Explanation:
## Step1: First, find the mass of the person \(m=\frac{W}{g}\), where \(W = 800\ N\) and \(g = 9.8\ m/s^{2}\), so \(m=\frac{800}{9.8}\ kg\).
Apply Newton's second law \(F_{net}=N - mg=ma\) (taking up as positive). Here \(a = 1\ m/s^{2}\) (upward because it is decelerating downward).
\(N=mg+ma=m(g + a)=\frac{800}{9.8}(9.8 + 1)\).
\(N=\frac{800\times10.8}{9.8}\approx877.6\ N\)
- # Answer:
Approximately 877.6 N
5. **For the 46 - gram golf ball shot straight up (speed 30 m/s, air - drag 0.14 N)**:
- **a) Imbalance?**
- # Explanation:
## Step1: There are two forces acting on the ball (gravity and air - drag) and the ball has an acceleration.
Since there is an acceleration, there is a net force, so there is an imbalance.
- # Answer:
Yes
- **b) Acceleration of the ball**:
- # Explanation:
## Step1: First, find the gravitational force \(F_g=mg\), where \(m = 0.046\ kg\) and \(g = 9.8\ m/s^{2}\), so \(F_g=0.046\times9.8 = 0.4508\ N\).
The net force \(F_{net}=F_g + F_{drag}\) (both forces act downward when the ball is moving up).
According to Newton's second law \(F_{net}=ma\), so \(a=\frac{F_g + F_{drag}}{m}\).
\(a=\frac{0.4508+0.14}{0.046}=\frac{0.5908}{0.046}\approx12.84\ m/s^{2}\) downward.
- # Answer:
Magnitude: approximately \(12.84\ m/s^{2}\), Direction: Downward
6. **For the 70 - kg physics student in the elevator (scale reads 660 N)**:
- **a) Imbalance?**
- # Explanation:
## Step1: The normal force (scale reading) is not equal to the gravitational force.
Since \(F_N eq mg\) (\(mg=70\times9.8 = 686\ N\) and \(F_N = 660\ N\)), there is a net force, so there is an imbalance.
- # Answer:
Yes
- **b) Acceleration of the elevator**:
- # Explanation:
## Step1: Apply Newton's second law \(F_{net}=mg - F_N=ma\).
Rearrange to find \(a=\frac{mg - F_N}{m}\).
Given \(m = 70\ kg\), \(mg = 686\ N\), and \(F_N = 660\ N\).
\(a=\frac{686 - 660}{70}=\frac{26}{70}\approx0.37\ m/s^{2}\) downward.
- # Answer:
Magnitude: approximately \(0.37\ m/s^{2}\), Direction: Downward
- **c) Can the elevator be moving upward?**
- # Explanation:
## Step1: The elevator can be moving upward while decelerating.
When the elevator is moving upward and decelerating, the normal force on the person is less than the gravitational force.
- # Answer:
Yes, if it is decelerating.
7. **For the 1400 - kg car parked on a hill (tilted 8 degrees)**:
- **a) Imbalance?**
- # Explanation:
## Step1: The car is at rest, but there is a component of the gravitational force acting down - the slope.
There must be a frictional force to balance this component. Since there is a non - zero component of force that needs to be balanced, there is an imbalance if we consider only the gravitational force components.
- # Answer:
Yes
- **b) Magnitude of the frictional force**:
- # Explanation:
## Step1: The frictional force \(F_f\) balances the component of the gravitational force along the slope \(F_{g\parallel}=mg\sin\theta\).
Given \(m = 1400\ kg\), \(g = 9.8\ m/s^{2}\), and \(\theta = 8^{\circ}\).
\(F_f=1400\times9.8\times\sin(8^{\circ})\).
\(F_f\approx1400\times9.8\times0.1392\approx1911.7\ N\)
- # Answer:
Approximately 1911.7 N
- **c) Minimum coefficient of static friction**:
- # Explanation:
## Step1: The maximum static - frictional force \(F_f=\mu_sN\), and \(N = mg\cos\theta\).
Since \(F_f = mg\sin\theta\), then \(\mu_s=\tan\theta\).
\(\mu_s=\tan(8^{\circ})\approx0.14\)
- # Answer:
Approximately 0.14

Answer:

1. For the car driving around a curve:

• a) Imbalance?
• # Explanation:

Step1: A car moving in a circular - path needs a centripetal force.

Since the car is moving in a curved path, there is a net force acting towards the center of the curve. So, there is an imbalance.

• # Answer:

Yes

1. For the skydiver at terminal velocity (80 - kg, 125 mph):

• a) Imbalance?
• # Explanation:

Step1: At terminal velocity, the net force on the skydiver is zero.

The gravitational force is balanced by the drag force. So, there is no imbalance.

• # Answer:

No

• b) Magnitude of the drag force:
• # Explanation:

Step1: At terminal velocity, \(F_{drag}=mg\).

Given \(m = 80\ kg\) and \(g = 9.8\ m/s^{2}\), then \(F_{drag}=80\times9.8\).
\(F_{drag}=784\ N\)

• # Answer:

784 N

1. For the 80 - kg skydiver falling at 40 m/s and accelerating at 2 m/s²:

• a) Imbalance?
• # Explanation:

Step1: Since there is an acceleration, there is a net force.

According to Newton's second law \(F = ma\), if \(a
eq0\), there is an imbalance.

• # Answer:

Yes

• b) Magnitude of the drag force:
• # Explanation:

Step1: Apply Newton's second law \(F_{net}=mg - F_{drag}=ma\).

Rearrange to find \(F_{drag}=mg - ma\).
Given \(m = 80\ kg\), \(g = 9.8\ m/s^{2}\), and \(a = 2\ m/s^{2}\).
\(F_{drag}=80\times9.8-80\times2\).
\(F_{drag}=80\times(9.8 - 2)=80\times7.8 = 624\ N\)

• # Answer:

624 N

1. For the person in the elevator (weighs 800 N, moving down at 4 m/s and slowing down at 1 m/s²):

• a) Imbalance?
• # Explanation:

Step1: Since the elevator is decelerating, there is a net force.

Deceleration implies a non - zero acceleration, so there is an imbalance.

• # Answer:

Yes

• b) Scale reading:
• # Explanation:

Step1: First, find the mass of the person \(m=\frac{W}{g}\), where \(W = 800\ N\) and \(g = 9.8\ m/s^{2}\), so \(m=\frac{800}{9.8}\ kg\).

Apply Newton's second law \(F_{net}=N - mg=ma\) (taking up as positive). Here \(a = 1\ m/s^{2}\) (upward because it is decelerating downward).
\(N=mg+ma=m(g + a)=\frac{800}{9.8}(9.8 + 1)\).
\(N=\frac{800\times10.8}{9.8}\approx877.6\ N\)

• # Answer:

Approximately 877.6 N

1. For the 46 - gram golf ball shot straight up (speed 30 m/s, air - drag 0.14 N):

• a) Imbalance?
• # Explanation:

Step1: There are two forces acting on the ball (gravity and air - drag) and the ball has an acceleration.

Since there is an acceleration, there is a net force, so there is an imbalance.

• # Answer:

Yes

• b) Acceleration of the ball:
• # Explanation:

Step1: First, find the gravitational force \(F_g=mg\), where \(m = 0.046\ kg\) and \(g = 9.8\ m/s^{2}\), so \(F_g=0.046\times9.8 = 0.4508\ N\).

The net force \(F_{net}=F_g + F_{drag}\) (both forces act downward when the ball is moving up).
According to Newton's second law \(F_{net}=ma\), so \(a=\frac{F_g + F_{drag}}{m}\).
\(a=\frac{0.4508+0.14}{0.046}=\frac{0.5908}{0.046}\approx12.84\ m/s^{2}\) downward.

• # Answer:

Magnitude: approximately \(12.84\ m/s^{2}\), Direction: Downward

1. For the 70 - kg physics student in the elevator (scale reads 660 N):

• a) Imbalance?
• # Explanation:

Step1: The normal force (scale reading) is not equal to the gravitational force.

Since \(F_N
eq mg\) (\(mg=70\times9.8 = 686\ N\) and \(F_N = 660\ N\)), there is a net force, so there is an imbalance.

• # Answer:

Yes

• b) Acceleration of the elevator:
• # Explanation:

Step1: Apply Newton's second law \(F_{net}=mg - F_N=ma\).

Rearrange to find \(a=\frac{mg - F_N}{m}\).
Given \(m = 70\ kg\), \(mg = 686\ N\), and \(F_N = 660\ N\).
\(a=\frac{686 - 660}{70}=\frac{26}{70}\approx0.37\ m/s^{2}\) downward.

• # Answer:

Magnitude: approximately \(0.37\ m/s^{2}\), Direction: Downward

• c) Can the elevator be moving upward?
• # Explanation:

Step1: The elevator can be moving upward while decelerating.

When the elevator is moving upward and decelerating, the normal force on the person is less than the gravitational force.

• # Answer:

Yes, if it is decelerating.

1. For the 1400 - kg car parked on a hill (tilted 8 degrees):

• a) Imbalance?
• # Explanation:

Step1: The car is at rest, but there is a component of the gravitational force acting down - the slope.

There must be a frictional force to balance this component. Since there is a non - zero component of force that needs to be balanced, there is an imbalance if we consider only the gravitational force components.

• # Answer:

Yes

• b) Magnitude of the frictional force:
• # Explanation:

Step1: The frictional force \(F_f\) balances the component of the gravitational force along the slope \(F_{g\parallel}=mg\sin\theta\).

Given \(m = 1400\ kg\), \(g = 9.8\ m/s^{2}\), and \(\theta = 8^{\circ}\).
\(F_f=1400\times9.8\times\sin(8^{\circ})\).
\(F_f\approx1400\times9.8\times0.1392\approx1911.7\ N\)

• # Answer:

Approximately 1911.7 N

• c) Minimum coefficient of static friction:
• # Explanation:

Step1: The maximum static - frictional force \(F_f=\mu_sN\), and \(N = mg\cos\theta\).

Since \(F_f = mg\sin\theta\), then \(\mu_s=\tan\theta\).
\(\mu_s=\tan(8^{\circ})\approx0.14\)

• # Answer:

Approximately 0.14