Question
Grade 12 Physics - Assignment 1 Newton’s Laws of Motion
Full Solution Assignments (40%) require thorough solutions and calculations presented in the GRASS format, which stands for Given, Required, Analysis, Solution, and Statement. Please refer to my notes for a detailed explanation of this format. Additionally, please note that all Dynamics problems must include a Free Body Diagram (FBD).
Problem:
A gardener pushes down along the handle of a lawn mower of 30 kg mass with a force of 180 N. The handle makes an angle of 42° with the ground. Calculate the instantaneous acceleration of the mower if the coefficient of kinetic frictions between its wheels and the ground at that instant is 0.62
Full Solution Assignments (40%) require thorough solutions and calculations presented in the GRASS format, which stands for Given, Required, Analysis, Solution, and Statement. Please refer to my notes for a detailed explanation of this format. Additionally, please note that all Dynamics problems must include a Free Body Diagram (FBD).
Problem:
A gardener pushes down along the handle of a lawn mower of 30 kg mass with a force of 180 N. The handle makes an angle of 42° with the ground. Calculate the instantaneous acceleration of the mower if the coefficient of kinetic frictions between its wheels and the ground at that instant is 0.62
Answers
GPT-4o mini
## GRASS Format Solution
### Given
- Mass of the lawn mower, \( m = 30 \, \text{kg} \)
- Force applied by the gardener, \( F = 180 \, \text{N} \)
- Angle of the handle with the ground, \( \theta = 42^\circ \)
- Coefficient of kinetic friction, \( \mu_k = 0.62 \)
### Required
- Instantaneous acceleration of the mower, \( a \)
### Analysis
1. **Free Body Diagram (FBD)**:
- Identify the forces acting on the lawn mower:
- Gravitational force \( F_g = mg \) acting downward.
- Normal force \( N \) acting upward.
- Applied force \( F \) which has two components:
- Horizontal component \( F_x = F \cos(\theta) \)
- Vertical component \( F_y = F \sin(\theta) \)
- Frictional force \( f_k = \mu_k N \) acting opposite to the direction of motion.
2. **Force Analysis**:
- Vertical forces:
\[
N + F_y = F_g \implies N + F \sin(\theta) = mg
\]
- Horizontal forces:
\[
F_x - f_k = ma
\]
3. **Calculations**:
- Gravitational force:
\[
F_g = mg = 30 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 294 \, \text{N}
\]
- Components of the applied force:
\[
F_x = F \cos(\theta) = 180 \cos(42^\circ)
\]
\[
F_y = F \sin(\theta) = 180 \sin(42^\circ)
\]
- Calculate \( F_x \) and \( F_y \):
- Using \( \cos(42^\circ) \approx 0.7431 \) and \( \sin(42^\circ) \approx 0.6691 \):
\[
F_x \approx 180 \times 0.7431 \approx 133.75 \, \text{N}
\]
\[
F_y \approx 180 \times 0.6691 \approx 120.43 \, \text{N}
\]
- Substitute \( F_y \) in the vertical force equation to find the normal force \( N \):
\[
N + 120.43 \, \text{N} = 294 \, \text{N} \implies N = 294 - 120.43 \approx 173.57 \, \text{N}
\]
- Find the frictional force:
\[
f_k = \mu_k N = 0.62 \times 173.57 \approx 107.76 \, \text{N}
\]
- Substitute values into the horizontal force equation:
\[
F_x - f_k = ma \implies 133.75 \, \text{N} - 107.76 \, \text{N} = 30 \, a
\]
\[
25.99 \, \text{N} = 30 \, a \implies a = \frac{25.99}{30} \approx 0.866 \, \text{m/s}^2
\]
### Solution
The instantaneous acceleration of the mower is approximately:
\[
\boxed{0.87 \, \text{m/s}^2}
\]
### Statement
The calculation shows that when a gardener pushes down on the handle of a 30 kg lawn mower, the instantaneous acceleration of the mower, accounting for friction and the force applied, is approximately \( 0.87 \, \text{m/s}^2 \).
### Given
- Mass of the lawn mower, \( m = 30 \, \text{kg} \)
- Force applied by the gardener, \( F = 180 \, \text{N} \)
- Angle of the handle with the ground, \( \theta = 42^\circ \)
- Coefficient of kinetic friction, \( \mu_k = 0.62 \)
### Required
- Instantaneous acceleration of the mower, \( a \)
### Analysis
1. **Free Body Diagram (FBD)**:
- Identify the forces acting on the lawn mower:
- Gravitational force \( F_g = mg \) acting downward.
- Normal force \( N \) acting upward.
- Applied force \( F \) which has two components:
- Horizontal component \( F_x = F \cos(\theta) \)
- Vertical component \( F_y = F \sin(\theta) \)
- Frictional force \( f_k = \mu_k N \) acting opposite to the direction of motion.
2. **Force Analysis**:
- Vertical forces:
\[
N + F_y = F_g \implies N + F \sin(\theta) = mg
\]
- Horizontal forces:
\[
F_x - f_k = ma
\]
3. **Calculations**:
- Gravitational force:
\[
F_g = mg = 30 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 294 \, \text{N}
\]
- Components of the applied force:
\[
F_x = F \cos(\theta) = 180 \cos(42^\circ)
\]
\[
F_y = F \sin(\theta) = 180 \sin(42^\circ)
\]
- Calculate \( F_x \) and \( F_y \):
- Using \( \cos(42^\circ) \approx 0.7431 \) and \( \sin(42^\circ) \approx 0.6691 \):
\[
F_x \approx 180 \times 0.7431 \approx 133.75 \, \text{N}
\]
\[
F_y \approx 180 \times 0.6691 \approx 120.43 \, \text{N}
\]
- Substitute \( F_y \) in the vertical force equation to find the normal force \( N \):
\[
N + 120.43 \, \text{N} = 294 \, \text{N} \implies N = 294 - 120.43 \approx 173.57 \, \text{N}
\]
- Find the frictional force:
\[
f_k = \mu_k N = 0.62 \times 173.57 \approx 107.76 \, \text{N}
\]
- Substitute values into the horizontal force equation:
\[
F_x - f_k = ma \implies 133.75 \, \text{N} - 107.76 \, \text{N} = 30 \, a
\]
\[
25.99 \, \text{N} = 30 \, a \implies a = \frac{25.99}{30} \approx 0.866 \, \text{m/s}^2
\]
### Solution
The instantaneous acceleration of the mower is approximately:
\[
\boxed{0.87 \, \text{m/s}^2}
\]
### Statement
The calculation shows that when a gardener pushes down on the handle of a 30 kg lawn mower, the instantaneous acceleration of the mower, accounting for friction and the force applied, is approximately \( 0.87 \, \text{m/s}^2 \).