Question

SPH4U
EXAM Part 1 – Important Submission Guidelines
(Weight: 20% of Course Grade)
EXAM Part 1 requires comprehensive solutions and calculations presented using the GRASS format, which stands for Given, Required, Analysis, Solution, and Statement.
For a detailed explanation of this format, please refer to my lecture notes. Please note that all Dynamics problems must include a clearly labeled Free Body Diagram (FBD) as part of your solution.
Problem:
A cart with a mass of 1.5 kg is pulled across a level desk by a force of 4.0 N at 8° above the horizontal. If the coefficient of kinetic friction is 0.12, what is the acceleration of the cart?

Answer 1

To solve the problem using the GRASS format, let's break it down as follows:

### Given:
- Mass of the cart, \( m = 1.5 \, \text{kg} \)
- Applied force, \( F = 4.0 \, \text{N} \)
- Angle of the applied force, \( \theta = 8° \)
- Coefficient of kinetic friction, \( \mu_k = 0.12 \)

### Required:
- Acceleration of the cart, \( a \)

### Analysis:
1. **Free Body Diagram (FBD)**:
- The forces acting on the cart include:
- The gravitational force (\( F_g = m \cdot g \))
- The normal force (\( N \))
- The applied force (\( F \)) which has a horizontal (\( F_x \)) and a vertical component (\( F_y \))
- The frictional force (\( F_f \))

The components of the applied force can be calculated as follows:
- \( F_x = F \cdot \cos(\theta) = 4.0 \cdot \cos(8°) \)
- \( F_y = F \cdot \sin(\theta) = 4.0 \cdot \sin(8°) \)

2. **Calculating Forces**:
- The weight of the cart:
\[
F_g = m \cdot g = 1.5 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 14.715 \, \text{N}
\]
- The normal force can be determined using the equilibrium in the vertical direction:
\[
N + F_y = F_g \implies N = F_g - F_y
\]

3. **Frictional Force**:
- The kinetic frictional force is given by:
\[
F_f = \mu_k \cdot N
\]

4. **Net Force and Acceleration**:
- Using Newton's second law:
\[
\Sigma F_x = F_x - F_f = m \cdot a
\]
- Rearrange to find acceleration \( a \).

### Solution:
1. **Calculate Components of the Applied Force**:
\[
F_x = 4.0 \cdot \cos(8°) \approx 4.0 \cdot 0.9903 \approx 3.9612 \, \text{N}
\]
\[
F_y = 4.0 \cdot \sin(8°) \approx 4.0 \cdot 0.1392 \approx 0.5568 \, \text{N}
\]

2. **Calculate Normal Force**:
\[
N = 14.715 \, \text{N} - 0.5568 \, \text{N} \approx 14.1582 \, \text{N}
\]

3. **Calculate Frictional Force**:
\[
F_f = 0.12 \cdot 14.1582 \, \text{N} \approx 1.698984 \, \text{N}
\]

4. **Calculate Net Force**:
\[
\Sigma F_x = F_x - F_f = 3.9612 \, \text{N} - 1.698984 \, \text{N} \approx 2.262216 \, \text{N}
\]

5. **Calculate Acceleration**:
\[
a = \frac{\Sigma F_x}{m} = \frac{2.262216 \, \text{N}}{1.5 \, \text{kg}} \approx 1.5081 \, \text{m/s}^2
\]

### Statement:
The acceleration of the cart is approximately \( 1.51 \, \text{m/s}^2 \).