A block of mass 0.8 kg is dragged along a level surface at constant velocity by a hanging block of mass 0.2 kg as shown in fig. 2.29. Find tension in the string and acceleration of the system, if the coefficient of friction between the block and the surface is 0.045 with diagram.
even i have the same query
I have the same quary
I'm sorry, but I don't have the ability to display or interpret diagrams. However, I can still help you calculate the tension in the string and the acceleration of the system.
To solve this problem, we can start by drawing a free-body diagram for each block. Let's designate the hanging block as Block A and the dragged block as Block B.
For Block A (hanging block):
- There is tension force (T) acting upwards.
- There is the force of gravity (m_A * g) acting downwards.
For Block B (dragged block):
- There is the force of tension (T) acting to the right.
- There is the force of friction (f) acting to the left.
- There is the force of gravity (m_B * g) acting downwards.
Since the blocks are moving at a constant velocity, the acceleration of the system is zero. This means that the net force acting on each block must be zero.
For Block A:
- The tension force (T) is equal to the force of gravity: T = m_A * g
For Block B:
- The force of tension (T) is equal to the force of friction: T = f
- The force of friction (f) can be calculated using the equation: f = μ * N, where μ is the coefficient of friction and N is the normal force.
- The normal force (N) is equal to the force of gravity: N = m_B * g
- Substituting the values, we get: f = μ * (m_B * g)
Since the net force on Block B is zero, we equate the force of tension and the force of friction:
T = f
μ * (m_B * g) = T
Now we can substitute the values and solve for T:
T = 0.045 * (0.2 * 9.8)
Calculating the tension:
T = 0.045 * 1.96
T ≈ 0.0882 N
So, the tension in the string is approximately 0.0882 N.
Since the system has zero acceleration, the net force on Block B is zero. This means that the force of tension (T) must be equal to the force of friction (f).
To find the acceleration, we can use Newton's second law:
Net force = mass * acceleration
For Block B:
T - f = m_B * a
Substituting the values:
(0.045 * (0.2 * 9.8)) - f = 0.8 * a
Simplifying the equation:
0.0882 - 0.045 * (0.2 * 9.8) = 0.8 * a
Solving for the acceleration:
0.0882 - 0.045 * 1.96 = 0.8 * a
a ≈ 0 m/s^2
Therefore, the acceleration of the system is approximately 0 m/s^2.
To find the tension in the string and the acceleration of the system, we can start by applying Newton's second law of motion. This law states that the net force on an object is equal to the mass of the object multiplied by its acceleration.
Let's break down the forces acting on the system:
1. Tension in the string: The tension in the string is the force exerted by the hanging block on the system. We'll denote this as T.
2. Force due to gravity: The hanging block has a mass of 0.2 kg and experiences a downward force due to gravity (weight) which can be calculated as Fg = m * g, where g is the acceleration due to gravity (approximated as 9.8 m/s^2).
3. Force due to friction: The block on the surface experiences a force opposing its motion, which is the force of friction. This force can be calculated as Ff = μ * N, where μ is the coefficient of friction and N is the normal force. In this case, since the block is on a level surface, the normal force is equal to the weight of the block, N = m * g.
Now, let's analyze the system:
Since the block is dragged at a constant velocity, the net force on the system is zero. This means that the tension in the string and the force of friction will cancel out the force due to gravity. Therefore, we can set up the following equation:
T - Ff = 0
Substituting the values for Ff and N, we have:
T - (μ * N) = 0
T - (μ * m * g) = 0
Now we can calculate the value of T:
T = μ * m * g
Given that μ = 0.045, m = 0.8 kg, and g = 9.8 m/s^2, we can substitute these values into the equation to find:
T = 0.045 * 0.8 kg * 9.8 m/s^2
T = 0.3528 N
So, the tension in the string is approximately 0.3528 N.
Now, to find the acceleration of the system, we can use the equation:
T - Ff = m * a
Substituting the values we know, we have:
0.3528 N - (0.045 * 0.8 kg * 9.8 m/s^2) = (0.8 kg + 0.2 kg) * a
Simplifying the equation, we get:
0.3528 N - 0.3528 N = 1 kg * a
0 = 1 kg * a
a = 0 m/s^2
So, the acceleration of the system is 0 m/s^2.
In conclusion, the tension in the string is approximately 0.3528 N, and the acceleration of the system is 0 m/s^2.