To solve this problem, we'll go step by step, using physics concepts related to work and energy.
Given Data:
- Mass of the block, \( m = 5 , \text{kg} \)
- Distance slid down the incline, \( d = 2.50 , \text{m} \)
- Angle of incline, \( \theta = 30^\circ \)
- Coefficient of kinetic friction, \( \mu_k = 0.40 \)
1. Work Done by the Force of Gravity
The gravitational force acts downwards, but when the block slides down the incline, we will compute the component of the gravitational force that acts along the incline.
- The weight of the block is given by \( W = m \cdot g = 5 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 49.05 , \text{N} \).
- The force due to gravity acting along the incline is \( F_{\text{gravity}} = W \cdot \sin(\theta) = m \cdot g \cdot \sin(30^\circ) \).
Calculating this: \[ F_{\text{gravity}} = 49.05 , \text{N} \cdot \sin(30^\circ) = 49.05 , \text{N} \cdot 0.5 = 24.525 , \text{N} \]
Now, the work done by the force of gravity as the block slides down the incline is given by: \[ W_{\text{gravity}} = F_{\text{gravity}} \cdot d = 24.525 , \text{N} \cdot 2.50 , \text{m} = 61.3125 , \text{J} \] Work done by gravity: \( \approx 61.31 , \text{J} \)
2. Work Done by the Friction Force
The frictional force opposes the motion of the block. The normal force acting on the block, \( F_N \), can be calculated as follows: \[ F_N = W \cdot \cos(\theta) = m \cdot g \cdot \cos(30^\circ) \]
Calculating: \[ F_N = 49.05 , \text{N} \cdot \cos(30^\circ) = 49.05 , \text{N} \cdot \frac{\sqrt{3}}{2} \approx 49.05 , \text{N} \cdot 0.866 = 42.5 , \text{N} \]
The frictional force \( F_{\text{friction}} \) is given by: \[ F_{\text{friction}} = \mu_k \cdot F_N = 0.40 \cdot 42.5 , \text{N} = 17.0 , \text{N} \]
The work done by the friction force is: \[ W_{\text{friction}} = - F_{\text{friction}} \cdot d = - 17.0 , \text{N} \cdot 2.50 , \text{m} = -42.5 , \text{J} \] Work done by friction: \( \approx -42.5 , \text{J} \)
3. Final Velocity at the Bottom of the Incline
The net work done on the block will determine its final kinetic energy. The net work \( W_{\text{net}} \) is: \[ W_{\text{net}} = W_{\text{gravity}} + W_{\text{friction}} = 61.3125 , \text{J} - 42.5 , \text{J} = 18.8125 , \text{J} \]
The work-energy principle states that the net work done is equal to the change in kinetic energy: \[ W_{\text{net}} = \Delta KE = KE_f - KE_i \] Since the block starts from rest, \( KE_i = 0 \), so: \[ W_{\text{net}} = KE_f = \frac{1}{2} m v^2 \] Setting the expressions equal: \[ 18.8125 , \text{J} = \frac{1}{2} \cdot 5 , \text{kg} \cdot v^2 \]
Solving for \( v \): \[ 18.8125 = \frac{5}{2} v^2 \] \[ 18.8125 = 2.5 v^2 \] \[ v^2 = \frac{18.8125}{2.5} = 7.525 \] \[ v = \sqrt{7.525} \approx 2.744 , \text{m/s} \]
Final velocity at the bottom of the incline: \( \approx 2.74 , \text{m/s} \)
Summary:
- Work done by the force of gravity: 61.31 J
- Work done by the friction force: -42.5 J
- Final velocity at the bottom of the incline: 2.74 m/s