In this problem, a 1 kg block starts sliding down a frictional inclined plane. The goal is to find the rate at which mechanical energy dissipates due to friction at \( t = 1 \) second. Let’s break this problem into steps and calculate the dissipated energy.
Given Data
- Mass of the block (m): 1 kg
- Inclination angle (θ): 37°
- Coefficient of friction (μ): 0.5
- Acceleration due to gravity (g): 9.8 m/s²
- Time (t): 1 second
Step 1: Forces Acting on the Block
There are two primary forces acting on the block as it slides down the incline: 1. The gravitational force, which has a component parallel to the incline causing the block to accelerate. 2. The frictional force, which opposes the motion of the block.
The gravitational force parallel to the incline is given by:
F_{\text{gravity}} = m g \sin(\theta)
Substituting the values:
F_{\text{gravity}} = 1 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \sin(37°)
Using $\sin(37°) \approx 0.6018$:
F_{\text{gravity}} ≈ 1 \times 9.8 \times 0.6018 ≈ 5.9 \, \text{N}
Next, we calculate the frictional force:
F_{\text{friction}} = \mu N
Where $N$ is the normal force, given by:
N = m g \cos(\theta)
Substituting values:
N = 1 \times 9.8 \times \cos(37°)
Using $\cos(37°) \approx 0.7986$:
N ≈ 1 \times 9.8 \times 0.7986 ≈ 7.8 \, \text{N}
Now, the frictional force becomes:
F_{\text{friction}} = 0.5 \times 7.8 \, \text{N} = 3.9 \, \text{N}
Step 2: Net Force and Acceleration
The net force \( F_{\text{net}} \) acting on the block is the difference between the gravitational force and the frictional force:
F_{\text{net}} = F_{\text{gravity}} - F_{\text{friction}} = 5.9 \, \text{N} - 3.9 \, \text{N} = 2.0 \, \text{N}
Using Newton’s second law $F = ma$, the acceleration of the block is:
a = \frac{F_{\text{net}}}{m} = \frac{2.0 \, \text{N}}{1 \, \text{kg}} = 2.0 \, \text{m/s}^2
Step 3: Velocity of the Block at \( t = 1 \) second
Since the block starts from rest, we can use the kinematic equation to find the velocity at \( t = 1 \) second:
v = u + at Where:
- u: initial velocity (0 m/s, since the block starts from rest),
- a: acceleration (2.0 m/s²),
- t: time (1 second).
Substituting values:
v = 0 + 2.0 \, \text{m/s}^2 \times 1 \, \text{s} = 2.0 \, \text{m/s}
Step 4: Rate of Energy Dissipation due to Friction
The rate at which mechanical energy is dissipated due to friction is the power dissipated by the frictional force. The power is given by:
P_{\text{friction}} = F_{\text{friction}} \times v
Substituting the values:
P_{\text{friction}} = 3.9 \, \text{N} \times 2.0 \, \text{m/s} = 7.8 \, \text{W}
Final Answer
The rate at which mechanical energy dissipates due to friction at \( t = 1 \) second is 7.8 W (watts).
