To tackle this problem, we need to analyze the forces acting on the box placed on the conveyor belt, which is inclined at a 30-degree angle. The box is moving relative to the belt, and we need to determine various aspects of its motion, including its acceleration, distance traveled before coming to rest, and the net force acting on it. Let's break this down step by step.
Understanding the Forces at Play
When the box is placed on the conveyor belt, several forces come into play:
- Gravitational Force (Weight): This acts vertically downward and can be calculated as \( W = mg \), where \( m \) is the mass of the box and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
- Normal Force: This acts perpendicular to the surface of the conveyor belt. Since the belt is inclined, the normal force will be less than the weight of the box.
- Frictional Force: This opposes the motion of the box relative to the belt. The maximum static friction can be calculated using \( f_{\text{max}} = \mu N \), where \( \mu \) is the coefficient of friction (0.8 in this case) and \( N \) is the normal force.
1. Relative Velocity of the Box
The box has a relative velocity of 3 m/s down the belt. This means that while the conveyor belt moves up at 3 m/s, the box is sliding down at the same speed. Therefore, the relative velocity of the box with respect to the belt is indeed 3 m/s down the belt.
2. Acceleration of the Box
To find the acceleration of the box, we need to consider the net force acting on it. The gravitational force can be resolved into two components: one parallel to the incline and one perpendicular to it.
- The component of gravitational force acting down the incline is given by \( W_{\parallel} = mg \sin(30^\circ) = \frac{mg}{2} \).
- The normal force can be calculated as \( N = mg \cos(30^\circ) = mg \frac{\sqrt{3}}{2} \).
- The frictional force opposing the motion is \( f = \mu N = 0.8 \times mg \frac{\sqrt{3}}{2} \).
The net force acting on the box can be expressed as:
Net Force = Gravitational Force Down the Incline - Frictional Force
Substituting the values, we get:
Net Force = \( \frac{mg}{2} - 0.8 \times mg \frac{\sqrt{3}}{2} \)
Now, using Newton's second law \( F = ma \), we can find the acceleration \( a \) of the box:
\( ma = \frac{mg}{2} - 0.8 \times mg \frac{\sqrt{3}}{2} \)
After simplifying, we find that the acceleration \( a \) is approximately \( 2 \, \text{m/s}^2 \) down the incline.
3. Distance Moved by the Box Before Coming to Rest
To find the distance the box travels before coming to rest, we can use the kinematic equation:
\( v^2 = u^2 + 2as \)
Where:
- \( v \) is the final velocity (0 m/s when it comes to rest),
- \( u \) is the initial velocity (3 m/s down the belt),
- \( a \) is the acceleration (which we found to be approximately \( -2 \, \text{m/s}^2 \), since it is acting against the motion),
- \( s \) is the distance traveled.
Rearranging the equation gives:
\( 0 = (3)^2 + 2(-2)s \)
Solving for \( s \), we find that the distance moved by the box before coming to rest is approximately \( 3.4 \, \text{m} \).
4. Net Force Acting on the Box
The net force acting on the box is directed down the belt. This is because the gravitational component acting down the incline exceeds the frictional force acting up the incline. Thus, the box accelerates down the belt until it comes to rest.
In summary, we have analyzed the motion of the box on the conveyor belt, calculated its acceleration, the distance it travels before stopping, and confirmed the direction of the net force acting on it. This comprehensive approach helps us understand the dynamics involved in such scenarios.
