QUESTION 3 A sled sits at the top of an icy hill that is 20 m tall. What is (the magnitude of) the velocity of the sled at the bottom of the hill? Assume there's no friction between the sled and the hill, and that the sled starts from rest.
QUESTION 3 A sled sits at the top of an icy hill that is 20 m tall. What is (the magnitude of) the velocity of the sled at the bottom of the hill? Assume there's no friction between the sled and the hill, and that the sled starts from rest.
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Question 3
#### Physics Problem: Calculating Velocity of a Sled
A sled sits at the top of an icy hill that is 20 m tall. What is the magnitude of the velocity of the sled at the bottom of the hill? Assume there is no friction between the sled and the hill, and that the sled starts from rest.
**Answer:**
---
#### Explanation:
To find the velocity of the sled at the bottom of the hill, we can utilize the principle of conservation of energy. The potential energy at the top will convert into kinetic energy at the bottom.
- **Potential Energy (PE) at the top:**
\[ PE = mgh \]
where \( m \) is the mass of the sled, \( g \) is the acceleration due to gravity (9.81 m/s²), and \( h \) is the height of the hill (20 m).
- **Kinetic Energy (KE) at the bottom:**
\[ KE = \frac{1}{2}mv^2 \]
where \( v \) is the velocity of the sled.
Since \( PE \) at the top converts entirely to \( KE \) at the bottom:
\[ mgh = \frac{1}{2}mv^2 \]
The mass \( m \) cancels out from both sides:
\[ gh = \frac{1}{2}v^2 \]
Solving for \( v \):
\[ v^2 = 2gh \]
\[ v = \sqrt{2gh} \]
\[ v = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 20 \, \text{m}} \]
\[ v = \sqrt{392} \, \text{m}^2/\text{s}^2 \]
\[ v \approx 19.8 \, \text{m/s} \]
Therefore, the magnitude of the velocity of the sled at the bottom of the hill is approximately \( 19.8 \, \text{m/s} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5bc418ab-5051-450d-ba5a-1e906e0fb9c9%2F3115c7c6-bc3e-4448-87c5-96f7239020f4%2F9uycse_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 3
#### Physics Problem: Calculating Velocity of a Sled
A sled sits at the top of an icy hill that is 20 m tall. What is the magnitude of the velocity of the sled at the bottom of the hill? Assume there is no friction between the sled and the hill, and that the sled starts from rest.
**Answer:**
---
#### Explanation:
To find the velocity of the sled at the bottom of the hill, we can utilize the principle of conservation of energy. The potential energy at the top will convert into kinetic energy at the bottom.
- **Potential Energy (PE) at the top:**
\[ PE = mgh \]
where \( m \) is the mass of the sled, \( g \) is the acceleration due to gravity (9.81 m/s²), and \( h \) is the height of the hill (20 m).
- **Kinetic Energy (KE) at the bottom:**
\[ KE = \frac{1}{2}mv^2 \]
where \( v \) is the velocity of the sled.
Since \( PE \) at the top converts entirely to \( KE \) at the bottom:
\[ mgh = \frac{1}{2}mv^2 \]
The mass \( m \) cancels out from both sides:
\[ gh = \frac{1}{2}v^2 \]
Solving for \( v \):
\[ v^2 = 2gh \]
\[ v = \sqrt{2gh} \]
\[ v = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 20 \, \text{m}} \]
\[ v = \sqrt{392} \, \text{m}^2/\text{s}^2 \]
\[ v \approx 19.8 \, \text{m/s} \]
Therefore, the magnitude of the velocity of the sled at the bottom of the hill is approximately \( 19.8 \, \text{m/s} \).
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