Two blocks of masses 11 kg and 8.0 kg are connected together by a light red string and rest on a frictionless level surface. Attached to the 8-kg mass is a light blue string, which a person uses to pull both blocks horizontally. If the two-block system accelerates at 0.5 m/s2, what is the tension in the blue string attached to the 8-kg mass?
T2=2mgsin0
Two bodies of mass M1 = 5Kg and M2 = 3Kg, are stationed on an incline. What is acceleration of the system, if the coefficient of kinetic friction is 0.5? What is the tension in the string?
To find the tension in the blue string attached to the 8-kg mass, we need to consider the forces acting on the two-block system.
Let's assume that the tension in the red string connecting the two blocks is T1, and the tension in the blue string attached to the 8-kg mass is T2.
The only horizontal force acting on the system is the tension in the blue string (T2). As there is no friction, this force will cause the system to accelerate.
We can apply Newton's second law of motion to the 8-kg mass:
T2 - T1 = m2 * a
where m2 represents the mass of the 8-kg block and a represents the acceleration of the system, which is given as 0.5 m/s^2.
Rearranging the equation, we get:
T2 = T1 + m2 * a
Now, let's find the value of T1. The only vertical forces acting on the two-block system are the gravitational forces on each block.
The gravitational force on the 11-kg block (m1) is given by:
F1 = m1 * g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Similarly, the gravitational force on the 8-kg block (m2) is:
F2 = m2 * g
Since the two blocks are connected by the red string, the tension in the red string (T1) will counterbalance the gravitational forces acting downwards:
T1 = F1 + F2 = (m1 + m2) * g
Substituting this value of T1 back into the equation for T2, we have:
T2 = (m1 + m2) * g + m2 * a
Plugging in the given values:
m1 = 11 kg
m2 = 8 kg
a = 0.5 m/s^2
g = 9.8 m/s^2
T2 = (11 + 8) * 9.8 + 8 * 0.5
T2 = 19 * 9.8 + 4
T2 ≈ 186.2 + 4
T2 ≈ 190.2 N
Therefore, the tension in the blue string attached to the 8-kg mass is approximately 190.2 N.
To find the tension in the blue string attached to the 8-kg mass, we can follow these steps:
Step 1: Calculate the net force acting on the system.
The net force acting on the system can be calculated using Newton's second law of motion. The formula is given by:
F_net = m_total * a
Where:
- F_net is the net force
- m_total is the total mass of the system
- a is the acceleration
In this case, the total mass of the system is the sum of the masses of the two blocks, which is 11 kg + 8 kg = 19 kg. The given acceleration is 0.5 m/s^2. Plugging these values into the formula:
F_net = 19 kg * 0.5 m/s^2 = 9.5 N
Step 2: Calculate the tension in the blue string.
Since the system is connected by a light string, the tension in the blue string attached to the 8-kg mass is the same as the tension in the red string connecting the two blocks. Let's call this tension T.
To calculate T, we need to consider the forces acting on the 8-kg mass. There are two forces: the tension T and the gravitational force pulling the mass downward.
The gravitational force (weight) acting on the 8-kg mass can be calculated using the formula:
F_gravity = m * g
Where:
- F_gravity is the gravitational force
- m is the mass of the object
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
In this case, the mass of the 8-kg block is 8 kg. Plugging this value into the formula:
F_gravity = 8 kg * 9.8 m/s^2 = 78.4 N
Since the system is in equilibrium (accelerating at a constant rate of 0.5 m/s^2), the net force is equal to 0. Therefore, the tension in the blue string is equal to the gravitational force acting on the 8-kg mass:
T = F_gravity = 78.4 N
So, the tension in the blue string attached to the 8-kg mass is 78.4 N.