Question

SPH4U - EXAM Part 1 requires comprehensive solutions and calculations presented using the
GRASS format, which stands for Given, Required, Analysis, Solution, and Statement.

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 this problem using the GRASS format, we'll break down each component accordingly.

### Given
- Mass of the cart (m) = 1.5 kg
- Pulling force (F) = 4.0 N
- Angle of force above the horizontal (θ) = 8°
- Coefficient of kinetic friction (μ) = 0.12
- Acceleration due to gravity (g) = 9.81 m/s²

### Required
- The acceleration (a) of the cart.

### Analysis
1. **Determine the components of the pulling force:**
- The horizontal component (F_x) and the vertical component (F_y) of the pulling force can be found using trigonometric functions.

\[
F_x = F \cdot \cos(θ), \quad F_y = F \cdot \sin(θ)
\]

2. **Calculate the normal force (N):**
The normal force will be affected by the vertical component of the pulling force.

\[
N = mg - F_y
\]

3. **Calculate the frictional force (f_k):**
The frictional force opposing the motion can be calculated as:

\[
f_k = μ \cdot N
\]

4. **Use Newton’s second law to find acceleration (a):**
The net force acting on the cart in the horizontal direction will be the difference between the horizontal pulling force and the frictional force.

\[
F_{net} = F_x - f_k
\]

Then, apply Newton's second law:

\[
F_{net} = m \cdot a
\]

From which we can solve for acceleration (a):

\[
a = \frac{F_{net}}{m}
\]

### Solution
1. **Calculate the components of the 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 the normal force (N):**
\[
N = mg - F_y = (1.5 \cdot 9.81) - 0.5568 \approx 14.715 - 0.5568 \approx 14.1582 \, \text{N}
\]

3. **Calculate the frictional force (f_k):**
\[
f_k = μ \cdot N = 0.12 \cdot 14.1582 \approx 1.6970 \, \text{N}
\]

4. **Calculate the net force (F_net):**
\[
F_{net} = F_x - f_k = 3.9612 - 1.6970 \approx 2.2642 \, \text{N}
\]

5. **Calculate the acceleration (a):**
\[
a = \frac{F_{net}}{m} = \frac{2.2642}{1.5} \approx 1.5095 \, \text{m/s²}
\]

### Statement
The acceleration of the cart is approximately **1.51 m/s²**.