Three blocks are located on a horizontal table. The coefficient of kinetic friction between the blocks and the table is 0.276. They are connected by a massless cord, as shown in the figure below, and pulled to the right. The masses of the three blocks are = 9.0 kg, = 0.5 kg, and = 5.0 kg. The pulling force is equal to = 82.0 N. What is the tension ?
The three boxes are all connected.
The tension in which cord? You have not provided nor sufficiently described the figure.
I suspect that there is a cord between each block and you want the tension in each one. If so, write a free body equation for each block and solve the three equations in three unkowns.
To find the tension in the cord, we need to consider the forces acting on each of the blocks.
First, let's consider the 9.0 kg block. The only force acting on it in the horizontal direction is the tension in the cord (T). Since the block is being pulled to the right, we can write a force equation as:
T - friction = mass * acceleration
The frictional force can be calculated as the product of the coefficient of kinetic friction (μ) and the normal force (N). The normal force is equal to the weight of the block (mass * gravity), where gravity is approximately 9.8 m/s^2.
The frictional force can be calculated as:
friction = μ * N = μ * (mass * gravity)
Now, let's consider the 0.5 kg block. It experiences two forces: the tension (T) in the cord pulling it to the right and the tension (T) in the cord pulling it to the left. We can write a force equation as:
T - T = mass * acceleration
0 = 0.5 kg * acceleration
Since the 0.5 kg block is not accelerating in the horizontal direction, the tensions cancel each other out.
Lastly, let's consider the 5.0 kg block. It experiences two forces: the tension (T) in the cord pulling it to the right and the frictional force (friction) between the block and the table. We can write a force equation as:
T - friction = mass * acceleration
Now, we have two equations with two unknowns (T and acceleration):
1) T - μ * (mass * gravity) = 9.0 kg * acceleration
2) T = 5.0 kg * acceleration + μ * (5.0 kg * gravity)
We can solve these equations simultaneously to find the tension (T) and acceleration.
To find the acceleration, we subtract equation (2) from equation (1):
(T - μ * (mass * gravity)) - (5.0 kg * acceleration + μ * (5.0 kg * gravity)) = 9.0 kg * acceleration - 5.0 kg * acceleration
Simplifying, we get:
T - 5.0 kg * acceleration - μ * (5.0 kg * gravity) = 9.0 kg * acceleration - 5.0 kg * acceleration
T - 5.0 kg * acceleration - μ * (5.0 kg * gravity) = 4.0 kg * acceleration
Rearranging the equation, we have:
9.0 kg * acceleration - 5.0 kg * acceleration - 4.0 kg * acceleration = T - μ * (5.0 kg * gravity)
(9.0 kg - 5.0 kg - 4.0 kg) * acceleration = T - μ * (5.0 kg * gravity)
0.0 kg * acceleration = T - μ * (5.0 kg * gravity)
T - μ * (5.0 kg * gravity) = 0
Since the 0.5 kg block is not accelerating, the tension in the cord (T) is equal to the friction between the 5.0 kg block and the table, which is given by:
friction = μ * N = μ * (mass * gravity) = 0.276 * (5.0 kg * 9.8 m/s^2)
Therefore, the tension in the cord (T) is equal to:
T = 0.276 * (5.0 kg * 9.8 m/s^2) = 13.608 N