Question
An air pump does 5,600 J of work to launch a water bottle rocket into the air. If the air pump applies 150 N of force to the rocket at an angle of 45° to the ground, what is the horizontal distance the water bottle rocket travels? Round your answer to two significant figures.
17 × 101 m
5.3 × 101 m
1.1 × 102 m
5.9 × 102 m
17 × 101 m
5.3 × 101 m
1.1 × 102 m
5.9 × 102 m
Answers
GPT-4o mini
To find the horizontal distance that the water bottle rocket travels, we first need to determine the horizontal component of the force applied by the air pump.
Given:
- Work done (W) = 5600 J
- Force (F) = 150 N
- Angle (θ) = 45°
The horizontal component (F_horizontal) of the force can be calculated using trigonometry:
\[ F_{\text{horizontal}} = F \cdot \cos(\theta) \]
\[ F_{\text{horizontal}} = 150 \, \text{N} \cdot \cos(45^\circ) \]
\[ F_{\text{horizontal}} = 150 \, \text{N} \cdot \frac{\sqrt{2}}{2} \]
\[ F_{\text{horizontal}} \approx 150 \cdot 0.7071 \approx 106.07 \, \text{N} \]
Next, we can determine the distance (d) the rocket travels using the work-energy principle. The work done on the rocket is equal to the force times the distance in the direction of the force. We can express this as:
\[ W = F_{\text{horizontal}} \cdot d \]
We can rearrange this to find the distance:
\[ d = \frac{W}{F_{\text{horizontal}}} \]
\[ d = \frac{5600 \, \text{J}}{106.07 \, \text{N}} \]
\[ d \approx 52.8 \, \text{m} \]
Rounding to two significant figures:
\[ d \approx 53 \, \text{m} \]
Among the given options, this rounds to \( 5.3 \times 10^1 \, \text{m} \).
Thus, the horizontal distance the water bottle rocket travels is:
**5.3 × 10¹ m**.
Given:
- Work done (W) = 5600 J
- Force (F) = 150 N
- Angle (θ) = 45°
The horizontal component (F_horizontal) of the force can be calculated using trigonometry:
\[ F_{\text{horizontal}} = F \cdot \cos(\theta) \]
\[ F_{\text{horizontal}} = 150 \, \text{N} \cdot \cos(45^\circ) \]
\[ F_{\text{horizontal}} = 150 \, \text{N} \cdot \frac{\sqrt{2}}{2} \]
\[ F_{\text{horizontal}} \approx 150 \cdot 0.7071 \approx 106.07 \, \text{N} \]
Next, we can determine the distance (d) the rocket travels using the work-energy principle. The work done on the rocket is equal to the force times the distance in the direction of the force. We can express this as:
\[ W = F_{\text{horizontal}} \cdot d \]
We can rearrange this to find the distance:
\[ d = \frac{W}{F_{\text{horizontal}}} \]
\[ d = \frac{5600 \, \text{J}}{106.07 \, \text{N}} \]
\[ d \approx 52.8 \, \text{m} \]
Rounding to two significant figures:
\[ d \approx 53 \, \text{m} \]
Among the given options, this rounds to \( 5.3 \times 10^1 \, \text{m} \).
Thus, the horizontal distance the water bottle rocket travels is:
**5.3 × 10¹ m**.