A worker pushed a 23 kg block 8.4 m along a level floor at constant speed with a force directed `32^(@)` below the horizontal . If the coefficient of kinetic friction between block and floor was 0.20, what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system ?

Answer

To solve the problem step by step, we will break it down into parts (a) and (b) as per the question.

### Given Data:
- Mass of the block, \( m = 23 \, \text{kg} \)
- Distance moved, \( s = 8.4 \, \text{m} \)
- Angle of applied force, \( \theta = 32^\circ \)
- Coefficient of kinetic friction, \( \mu = 0.20 \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Part (a): Work Done by the Worker’s Force

1. **Calculate the weight of the block:**
\[ W = mg = 23 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 225.4 \, \text{N} \]

2. **Identify the components of the applied force:**
Let \( F \) be the force applied by the worker.
- Horizontal component: \( F \cos \theta \)
- Vertical component: \( F \sin \theta \)

3. **Determine the normal force \( N \):**
The normal force can be calculated using the equilibrium in the vertical direction:
\[ N = mg + F \sin \theta \]

4. **Calculate the frictional force \( f_k \):**
The frictional force is given by:
\[ f_k = \mu N = \mu (mg + F \sin \theta) \]

5. **Set up the equation for horizontal motion:**
Since the block is moving at constant speed, the net force in the horizontal direction is zero:
\[ F \cos \theta = f_k \]
Substituting for \( f_k \):
\[ F \cos \theta = \mu (mg + F \sin \theta) \]

6. **Rearranging the equation:**
\[ F \cos \theta - \mu F \sin \theta = \mu mg \]
\[ F (\cos \theta - \mu \sin \theta) = \mu mg \]
\[ F = \frac{\mu mg}{\cos \theta - \mu \sin \theta} \]

7. **Substituting values to find \( F \):**
\[ F = \frac{0.20 \times 23 \times 9.8}{\cos 32^\circ - 0.20 \sin 32^\circ} \]
- Calculate \( \cos 32^\circ \approx 0.848 \) and \( \sin 32^\circ \approx 0.529 \):
\[ F = \frac{0.20 \times 225.4}{0.848 - 0.20 \times 0.529} \]
\[ F = \frac{45.08}{0.848 - 0.1058} = \frac{45.08}{0.7422} \approx 60.8 \, \text{N} \]

8. **Calculate the work done by the worker’s force:**
The work done \( W \) is given by:
\[ W = F \cos \theta \cdot s \]
\[ W = 60.8 \cdot \cos 32^\circ \cdot 8.4 \]
\[ W = 60.8 \cdot 0.848 \cdot 8.4 \approx 460.2 \, \text{J} \]

### Part (b): Increase in Thermal Energy of the Block-Floor System

1. **Calculate the frictional force:**
Using the previously calculated \( N \):
\[ N = mg + F \sin \theta \]
\[ N = 225.4 + 60.8 \cdot \sin 32^\circ \]
\[ N = 225.4 + 60.8 \cdot 0.529 \approx 225.4 + 32.2 \approx 257.6 \, \text{N} \]
\[ f_k = \mu N = 0.20 \cdot 257.6 \approx 51.52 \, \text{N} \]

2. **Calculate the work done by friction:**
The work done by the friction force is:
\[ W_f = f_k \cdot s = 51.52 \cdot 8.4 \approx 432.77 \, \text{J} \]

3. **Increase in thermal energy:**
The work done against friction is equal to the increase in thermal energy:
\[ \text{Increase in thermal energy} = W_f \approx 432.77 \, \text{J} \]

### Final Answers:
(a) The work done by the worker's force is approximately **460.2 J**.
(b) The increase in thermal energy of the block-floor system is approximately **432.77 J**.