To solve this problem, we'll break it down into parts:
Given Data:
- Mass of the block, \( m = 5.00 , \text{kg} \)
- Distance along the incline, \( d = 2.50 , \text{m} \)
- Angle of the incline, \( \theta = 30.0^\circ \)
- Coefficient of kinetic friction, \( \mu_k = 0.40 \)
- Gravitational acceleration, \( g = 9.81 , \text{m/s}^2 \)
1. Work Done by the Force of Gravity
The work done by gravity can be determined using the formula: \[ W_{\text{gravity}} = mgh \] where \( h \) is the vertical height gained or lost, which can be calculated using: \[ h = d \sin(\theta) \]
For our values: \[ h = 2.50 , \text{m} \cdot \sin(30.0^\circ) = 2.50 , \text{m} \cdot 0.5 = 1.25 , \text{m} \]
Now substituting in the work done by gravity: \[ W_{\text{gravity}} = 5.00 , \text{kg} \cdot 9.81 , \text{m/s}^2 \cdot 1.25 , \text{m} = 61.3125 , \text{J} \] Therefore: \[ W_{\text{gravity}} \approx 61.31 , \text{J} \]
2. Work Done by the Friction Force
First, we need to calculate the normal force acting on the block, \( N \): \[ N = mg \cos(\theta) \]
Calculating \( N \): \[ N = 5.00 , \text{kg} \cdot 9.81 , \text{m/s}^2 \cdot \cos(30.0^\circ) \] \[ N \approx 5.00 \cdot 9.81 \cdot 0.866 \approx 42.44 , \text{N} \]
Now, the friction force \( f_k \) is given by: \[ f_k = \mu_k N = 0.40 \cdot 42.44 \approx 16.976 , \text{N} \]
The work done by the friction force is: \[ W_{\text{friction}} = -f_k d = -16.976 , \text{N} \cdot 2.50 , \text{m} \approx -42.44 , \text{J} \] Therefore: \[ W_{\text{friction}} \approx -42.44 , \text{J} \]
3. Final Velocity at the Bottom of the Incline
Using the work-energy principle, the total work done on the block equals the change in kinetic energy. The block starts from rest, so its initial kinetic energy \( KE_{\text{initial}} = 0 \).
The total work done on the block is: \[ W_{\text{total}} = W_{\text{gravity}} + W_{\text{friction}} = 61.31 , \text{J} - 42.44 , \text{J} = 18.87 , \text{J} \]
The work done is equal to the change in kinetic energy: \[ W_{\text{total}} = KE_{\text{final}} - KE_{\text{initial}} = KE_{\text{final}} \]
The kinetic energy is given by: \[ KE_{\text{final}} = \frac{1}{2} mv^2 \]
Setting the equations equal: \[ 18.87 , \text{J} = \frac{1}{2} \cdot 5.00 , \text{kg} \cdot v^2 \]
Solving for \( v^2 \): \[ v^2 = \frac{2 \cdot 18.87}{5.00} = \frac{37.74}{5.00} \approx 7.548 \]
Now, solving for \( v \): \[ v = \sqrt{7.548} \approx 2.74 , \text{m/s} \]
Summary of Results:
- Work done by gravity: \( W_{\text{gravity}} \approx 61.31 , \text{J} \)
- Work done by friction: \( W_{\text{friction}} \approx -42.44 , \text{J} \)
- Final Velocity at the bottom of the incline: \( v \approx 2.74 , \text{m/s} \)