Two blocks are connected by a (massless) string that passes over a (massless, frictionless) pulley as shown. Assume the interface between block A and the horizontal tabletop to be frictionless. If mA = 8 kg and mB = 8.7 kg, what is the tension in the string?Express your answer in N, to at least one digit after the decimal point.

To solve this problem, we need to use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

First, we need to calculate the acceleration of the system. The only forces acting on the system are the weight of block B pulling down and the weight of block A pulling up. The net force is the difference between these two forces.

The weight of an object is its mass times the acceleration due to gravity. So, the weight of block B is mBg and the weight of block A is mAg. The net force is mBg - mAg.

The acceleration of the system is the net force divided by the total mass, which is mA + mB. So, the acceleration is (mBg - mAg) / (mA + mB).

Given that mA = 8 kg, mB = 8.7 kg, and g (acceleration due to gravity) is approximately 9.8 m/s^2, we can substitute these values into the equation to find the acceleration:

a = (8.7 kg * 9.8 m/s^2 - 8 kg * 9.8 m/s^2) / (8 kg + 8.7 kg) = 0.98 m/s^2

Next, we need to calculate the tension in the string. The tension is the force that block A exerts on the string, which is equal to the weight of block A minus the force that block A exerts on the table. Since there is no friction between block A and the table, the force that block A exerts on the table is just its weight, which is mAg. So, the tension is mAg - mA*a.

Substituting the given values into this equation, we find:

T = 8 kg * 9.8 m/s^2 - 8 kg * 0.98 m/s^2 = 70.4 N

So, the tension in the string is 70.4 N.

To solve this problem, we need to use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

First, we need to calculate the acceleration of the system. The only forces acting on the system are the weight of block B pulling down and the weight of block A pulling up. The net force is the difference between these two forces.

The weight of an object is its mass times the acceleration due to gravity. So, the weight of block B is mBg and the weight of block A is mAg. The net force is mBg - mAg.

The acceleration of the system is the net force divided by the total mass of the system. So, the acceleration is (mBg - mAg) / (mA + mB).

Substituting the given values, we get:

Acceleration = (8.7 kg * 9.8 m/s^2 - 8 kg * 9.8 m/s^2) / (8 kg + 8.7 kg) = 0.98 m/s^2

Next, we calculate the tension in the string. The tension is the force that block A exerts on the string, which is equal to the force that the string exerts on block A.

According to Newton's second law, the force that the string exerts on block A is equal to the mass of block A times its acceleration. So, the tension is mA*a.

Substituting the given values, we get:

Tension = 8 kg * 0.98 m/s^2 = 7.84 N

So, the tension in the string is 7.84 N.

To solve this problem, we need to use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

First, we need to find the acceleration of the system. The only forces acting on the system are the weight of block B pulling down and the weight of block A pulling up. The net force is the difference between these two forces.

The weight of an object is its mass times the acceleration due to gravity. So, the weight of block B is mBg and the weight of block A is mAg. The net force is mBg - mAg.

The acceleration of the system is the net force divided by the total mass of the system. So, the acceleration is (mBg - mAg) / (mA + mB).

Substituting the given values, we get:

Acceleration = (8.7 kg * 9.8 m/s^2 - 8 kg * 9.8 m/s^2) / (8 kg + 8.7 kg) = 0.98 m/s^2

Next, we find the tension in the string. The tension is the force that block A exerts on the string, which is equal to the force that the string exerts on block A. This force is the weight of block A minus the force of block A on the string due to the acceleration of the system.

So, the tension is mAg - mAa.

Substituting the given values, we get:

Tension = 8 kg * 9.8 m/s^2 - 8 kg * 0.98 m/s^2 = 70.4 N

So, the tension in the string is 70.4 N.

To solve this problem, we need to use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

First, we need to find the acceleration of the system. The only forces acting on the system are the weight of block B pulling down and the weight of block A pulling up. The net force on the system is the difference between these two forces.

The weight of an object is its mass times the acceleration due to gravity. So, the weight of block B is mBg and the weight of block A is mAg. The net force on the system is mBg - mAg.

According to Newton's second law, the net force on the system is also equal to the total mass of the system times its acceleration. So, we can set up the equation mBg - mAg = (mA + mB)*a and solve for a.

Given that mA = 8 kg, mB = 8.7 kg, and g = 9.8 m/s^2, we find that a = (8.79.8 - 89.8) / (8 + 8.7) = 0.98 m/s^2.

Next, we find the tension in the string. The tension is the force that block A feels from the string, which is equal to the force required to move block A with the acceleration we found. According to Newton's second law, this force is mA*a.

So, the tension in the string is 8 kg * 0.98 m/s^2 = 7.84 N.