https://file.aiquickdraw.com/mathbot/file/1727376576152pi5fwdx0.jpg

To solve this problem, we need to calculate the acceleration of the block when \( x = 0.410 \, \text{m} \). Here's a breakdown of the forces involved and how to calculate the acceleration.

### Given dataTo solve this problem, we need to calculate the acceleration of the block when \( x = 0.410 \, \text{m} \). Here's a breakdown of the forces involved and how to calculate the acceleration.

### Given 
- Mass of the block, \( m = 2.02 \, \text{kg} \)
- Tension in the rope, \( T = 13.8 \, \text{N} \)
- Coefficient of kinetic friction, \( \mu_k = 0.375 \)
- Distance \( x = 0.410 \, \text{m} \)
- Pulley height above the block is \( h = 11.9 \, \text{cm} = 0.119 \, \text{m} \)

### Step 1: Normal Force and Friction
The block is on a horizontal surface, so the normal force is equal to the weight of the block. The frictional force is given by:
\[
f_k = \mu_k \cdot N = \mu_k \cdot m \cdot g
\]
where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity.

### Step 2: Force Analysis in the Horizontal Direction
We need to find the horizontal component of the tension force \( T_x \). The total tension \( T \) acts at an angle because the pulley is elevated by a height \( h \).

First, we calculate the angle \( \theta \) from the geometry of the problem:
\[
\theta = \tan^{-1}\left(\frac{h}{x}\right) = \tan^{-1}\left(\frac{0.119}{0.410}\right)
\]
Once we have \( \theta \), the horizontal component of the tension is:
\[
T_x = T \cdot \cos(\theta)
\]

### Step 3: Applying Newton's Second Law
The net force in the horizontal direction is the difference between the horizontal component of the tension and the kinetic friction force:
\[
F_{\text{net}} = T_x - f_k
\]
The acceleration of the block is then:
\[
a = \frac{F_{\text{net}}}{m}
\]

Let me calculate the values for the tension component, friction, and acceleration.The acceleration of the block is approximately \( 2.88 \, \text{m/s}^2 \), and the angle \( \theta \) at which the rope pulls the block is about \( 16.19^\circ \).

Would you like further details or explanations on any part of this solution?

Here are some related questions for further exploration:
1. How would the acceleration change if the coefficient of kinetic friction were higher?
2. What happens to the acceleration if the mass of the block increases?
3. How does the angle of the rope affect the horizontal component of the tension force?
4. What would be the effect of reducing the tension in the rope to 10 N?
5. How would the acceleration differ if the pulley were at a lower height?

**Tip:** The angle of a rope in pulley problems is crucial since it changes the effective tension in the horizontal direction. Always resolve forces into components!

A block of mass 2.02 kg is accelerated across a rough surface by a rope passing over a pulley. The tension in the rope is 13.8 N, and the pulley is 11.9 cm above the top of the block. The coefficient of kinetic friction is 0.375. Determine the acceleration of the block when x = 0.410 m.

Learn how to calculate the acceleration of a 2.02 kg block being pulled across a rough surface with a pulley system. This problem involves Newton's Laws of Motion, vector decomposition, and frictional forces. Tension in the rope is 13.8 N, with a friction coefficient of 0.375, and the pulley is 11.9 cm above the block. Suitable for advanced high school or college-level physics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation