A block of mass 6.0 kg is pushed up an incline to its top by a man and then allowed to slide down to the bottom. The length of incline is 10 m and its height is 5.0 m. the coefficient of friction between block and incline is 0.40. Calculate (a) the work done by the gravitational force over the complete round trip of the block, (b) the work done by the man during the upward journey, (c) the mechanical energy loss due to friction over the round trip, and (d) the speed of the block when it reaches the bottom.

Answer

To solve the problem step by step, we will break it down into the four parts as stated in the question.

### Given Data:
- Mass of the block, \( m = 6.0 \, \text{kg} \)
- Length of incline, \( L = 10 \, \text{m} \)
- Height of incline, \( h = 5.0 \, \text{m} \)
- Coefficient of friction, \( \mu = 0.40 \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)

### Step 1: Calculate the angle of incline (\( \theta \))
Using the sine function:
\[ \sin \theta = \frac{h}{L} = \frac{5}{10} = 0.5 \implies \theta = 30^\circ \]

### Part (a): Work done by the gravitational force over the complete round trip
1. **Work done while going up (A to B)**:
\[ W_{up} = -mg \sin \theta \cdot L \]
\[ W_{up} = -6 \cdot 9.8 \cdot \sin(30^\circ) \cdot 10 = -6 \cdot 9.8 \cdot 0.5 \cdot 10 = -294 \, \text{J} \]

2. **Work done while coming down (B to A)**:
\[ W_{down} = mg \sin \theta \cdot L \]
\[ W_{down} = 6 \cdot 9.8 \cdot \sin(30^\circ) \cdot 10 = 294 \, \text{J} \]

3. **Total work done by gravitational force**:
\[ W_{total} = W_{up} + W_{down} = -294 + 294 = 0 \, \text{J} \]

### Part (b): Work done by the man during the upward journey
1. **Calculate normal force (\( N \))**:
\[ N = mg \cos \theta = 6 \cdot 9.8 \cdot \cos(30^\circ) = 6 \cdot 9.8 \cdot \frac{\sqrt{3}}{2} \approx 50.9 \, \text{N} \]

2. **Calculate frictional force (\( F_k \))**:
\[ F_k = \mu N = 0.40 \cdot 50.9 \approx 20.36 \, \text{N} \]

3. **Calculate total force exerted by the man (\( F \))**:
\[ F = mg \sin \theta + F_k = 6 \cdot 9.8 \cdot 0.5 + 20.36 \approx 29.4 + 20.36 \approx 49.76 \, \text{N} \]

4. **Work done by the man**:
\[ W_{man} = F \cdot L = 49.76 \cdot 10 \approx 497.6 \, \text{J} \]

### Part (c): Mechanical energy loss due to friction over the round trip
1. **Work done against friction going up (A to B)**:
\[ W_{friction, up} = -F_k \cdot L = -20.36 \cdot 10 \approx -203.6 \, \text{J} \]

2. **Work done against friction coming down (B to A)**:
\[ W_{friction, down} = F_k \cdot L = 20.36 \cdot 10 \approx 203.6 \, \text{J} \]

3. **Total mechanical energy loss due to friction**:
\[ E_{loss} = W_{friction, up} + W_{friction, down} = -203.6 + 203.6 = -407.2 \, \text{J} \]

### Part (d): Speed of the block when it reaches the bottom
1. **Using the work-energy principle**:
\[ W_{gravity} + W_{friction} = \Delta KE \]
\[ -294 + (-203.6) = \frac{1}{2} mv^2 \]
\[ -497.6 = \frac{1}{2} \cdot 6 \cdot v^2 \]
\[ -497.6 = 3v^2 \implies v^2 = \frac{-497.6}{3} \implies v^2 = 165.87 \implies v \approx 12.87 \, \text{m/s} \]

### Summary of Results:
- (a) Work done by gravitational force: \( 0 \, \text{J} \)
- (b) Work done by the man: \( 497.6 \, \text{J} \)
- (c) Mechanical energy loss due to friction: \( 407.2 \, \text{J} \)
- (d) Speed of the block when it reaches the bottom: \( 12.87 \, \text{m/s} \)