01:24:00
NPH-161 ### 1. Understanding the Problem
Before diving into solving the problem, it's crucial to thoroughly understand what is being asked. Here's the problem statement:
> A skier sk on a hill that is uniformly sloped with a slope of 0.40. The friction between the skier and the hill is 0.23. The sk's acceleration is _________. (a) 1.24 m/s^2 (b) 0.1 m/s2 (c) 1.76 m/s2 (d) 0.13 m/s2
From this, I can extract that there's a skier on a hill with a slope of 0.40. There's friction between the skier and the hill, which is 0.23, and I need to find the acceleration of the skier.
### 2. Breaking Down the Problem
Next, I need to analyze the problem to understand what's given and what needs to be found.
- **Given:**
- Slope of the hill: 0.40
- Friction between skier and hill: 0.23
- Options for acceleration: (a) 1.24 m/s^2 (b) 0.1 m/s2 (c) 1.764 m/s2 (d) 0.13 m/s2
- **Need to find:**
- The acceleration of the skier
### 3. Knowing the Formulas
To solve this problem, I need to recall some physics principles, particularly Newton's second law of motion which states:
`F = m * a`
Where:
F is the total force on the object
m is the mass of the object
a is the acceleration of the object
Also, I need to know about the Normal force which is given by:
`N = m * g * cos(theta)`
Where:
N is the normal force
m is the mass of the object
g is the acceleration due to gravity (approximately 9.81 m/s^2)
theta is the angle of the hill
Since the slope is given as 0.40, I can calculate the angle of the hill using the formula:
`slope = tan(theta)`
So, theta = atan(slope) = atan(0.40) = 21.8 degrees
### 4. Calculate the Forces
There are forces acting on the skier:
1. The component of the object's weight that is parallel to the hill: This is given by:
`F_parallel = m * g * sin(theta)`
2. The friction force: This is given by:
`F_friction = m * g * cos(theta) * friction`
Since the skier is going downhill, the friction force is acting upwards, opposing the motion.
The net force on the skier is:
`F_net = F_parallel - F_friction`
### 5. Applying Newton's Second Law
To find the acceleration, I can rearrange the formula `F = m * a` to:
`a = F / m`
Since all the forces are expressed in terms of mass (m), I can eliminate m from the calculations.
So, the acceleration is given by:
`a = (F_parallel - F_friction) / m`
Since `F_parallel = m * g * sin(theta)` and `F_friction = m * g * cos(theta) * friction`, the equation becomes:
`a = (m * g * sin(theta) - m * g * cos(theta) * friction) / m`
`a = g * (sin(theta) - cos(theta) * friction)`
### 6. Calculating the Value
Now, I can substitute the known values into the equation:
`a = 9.81 * (sin(21.8) - cos(21.8) * 0.23)`
First, calculate sin(21.8) and cos(21.8):
`sin(21.8) = 0.371`
`cos(21.8) = 0.930`
`a = 9.81 * (0.371 - 0.930 * 0.23)`
`a = 9.81 * (0.371 - 0.2139)`
`a = 9.81 * 0.1571`
`a = 1.5429 m/s^2`
### 7. Choosing the Correct Answer
Now, compare this value with the given options:
(a) 1.24 m/s^2 (b) 0.1 m/s2 (c) 1.764 m/s2 (d) 0.13 m/s2
Since 1.5429 m/s^2 is closest to 1.764 m/s2, the correct answer is (c) 1.764 m/s2.
### 8. Final Answer
The correct answer is `(c) 1.764 m/s²`.
---
**Note**####
Here's a step-by-step breakdown of how to solve the problem:
1. **Understanding the Problem:**
- **Given:**
- Slope of the hill: 0.40
- Friction between skier and hill: 0.23
- Options for acceleration: (a) 1.24 m/s^2 (b) 0.1 m/s^2 (c) 1.76 m/s^2 (d) 0.13 m/s^2
- **Need to find:** The acceleration of the skier
2. **Breaking Down the Problem:**
- **Known:**
- Slope = 0.40
- Friction = 0.23
- Acceleration options: (a) 1.24 (b) 0.1 (c) 1.76 (d) 0.13
- **Unknown:**
- Acceleration of skier
3. **Knowing the Formulas:**
- Recall Newton's second law: `F = m * a`
- Forces on the skier:
- `F_parallel` = mg sin(theta)
- `F_friction` = m g cos(theta) * friction
- Net force: `F_net` = F_parallel - F_friction
- Acceleration: `a = F_net / m`
4. **Calculating Starting Values:**
- **Theta calculation:**
- `Slope` = tan(theta) = 0.40
- `theta` = atan(0.40) = 21.8 degrees
5. **Calculating the Forces:**
- `F_parallel = m * g * sin(theta) = m * 9.81 * sin(21.8) = m * 9.81 * 0.371 = m * 3.64`
- `F_friction = m * g * cos(theta) * friction = m * 9.81 * cos(21.8) * 0.23 = m * 9.81 * 0.930 * 0.23 = m * 2.04
- Net force: `F_net` = F_parallel - F_friction = m * 3.64 - m * 2.04 = m * 1.60
- Acceleration: `a = F_net / m`= m * 1.60 / m = 1.60 m/s^2
6. **Choosing the Correct Answer:**
- Closest option to 1.60 m/s^2 is (c) 1.76 m/s^2
7. **Final Answer:**
- The correct answer is `(c) 1.76 m/s²`
---
**Note**
28 Apr 2025
English
中文
日本語
Bahasa Indo
