A horizontal force of magnitude 41.0 N pushes a block of mass 4.00 kg across a floor where the coefficient of kinetic friction is 0.600. (a) How much work is done by that applied force on the block-floor system when the block slides through a displacement of 2.00 m across the floor ? (b) During that displacement, the thermal energy of the block increases by 40.0 J. What is the increase in thermal energy of the floor ? (c) What is the increase in the kinetic energy of the block ?

Answer

To solve the given problem step by step, we will break it down into three parts as per the question.

### Given Data:
- Mass of the block, \( m = 4.00 \, \text{kg} \)
- Applied force, \( F = 41.0 \, \text{N} \)
- Coefficient of kinetic friction, \( \mu_k = 0.600 \)
- Displacement, \( s = 2.00 \, \text{m} \)
- Increase in thermal energy of the block, \( \Delta E_{\text{block}} = 40.0 \, \text{J} \)

### Part (a): Work Done by the Applied Force
The work done \( W \) by the applied force can be calculated using the formula:
\[ W = F \cdot s \cdot \cos(\theta) \]
Since the force is applied horizontally and the displacement is also horizontal, the angle \( \theta = 0^\circ \) and \( \cos(0) = 1 \).

Substituting the values:
\[ W = 41.0 \, \text{N} \cdot 2.00 \, \text{m} \cdot 1 \]
\[ W = 82.0 \, \text{J} \]

### Part (b): Increase in Thermal Energy of the Floor
First, we need to calculate the work done by the friction force. The friction force \( F_{\text{friction}} \) can be calculated as:
\[ F_{\text{friction}} = \mu_k \cdot N \]
where \( N \) (the normal force) is equal to the weight of the block:
\[ N = m \cdot g = 4.00 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \approx 39.24 \, \text{N} \]
Now substituting the values:
\[ F_{\text{friction}} = 0.600 \cdot 39.24 \approx 23.54 \, \text{N} \]

The work done by the friction force over the displacement of 2 m is:
\[ W_{\text{friction}} = -F_{\text{friction}} \cdot s = -23.54 \, \text{N} \cdot 2.00 \, \text{m} \approx -47.08 \, \text{J} \]
(The negative sign indicates that the friction force opposes the motion.)

The total thermal energy generated during the displacement is equal to the work done by the friction force:
\[ \Delta E_{\text{total}} = -W_{\text{friction}} = 47.08 \, \text{J} \]

Given that the thermal energy of the block increased by 40.0 J, the increase in thermal energy of the floor can be calculated as:
\[ \Delta E_{\text{floor}} = \Delta E_{\text{total}} - \Delta E_{\text{block}} = 47.08 \, \text{J} - 40.0 \, \text{J} \approx 7.08 \, \text{J} \]

### Part (c): Increase in Kinetic Energy of the Block
The total work done by the applied force is equal to the increase in kinetic energy plus the thermal energy generated:
\[ W = \Delta KE + \Delta E_{\text{total}} \]
Rearranging gives:
\[ \Delta KE = W - \Delta E_{\text{total}} \]
Substituting the values:
\[ \Delta KE = 82.0 \, \text{J} - 47.08 \, \text{J} \approx 34.92 \, \text{J} \]

### Final Answers:
(a) Work done by the applied force: **82.0 J**
(b) Increase in thermal energy of the floor: **7.08 J**
(c) Increase in kinetic energy of the block: **34.92 J**