A blob of mass 5 kg is place on a table top.it is attached to a mass piece of 2kg by a light string that passes over a frictionless pulley.the table top exerts a frictional force of 7 N on the block (a) calculate the tension of the string

first get acceleration of the two masses

a = F/m = (2*9.81)/(7)

then:

mg - T = m a

2*9.81 - T = 2 a

T = 2 (9.81 - a)

To calculate the tension in the string, we need to consider the forces acting on each object.

1. For the 5 kg block:
- The weight of the block is given by: weight = mass * gravity.
- Weight = 5 kg * 9.8 m/s^2 = 49 N (upwards).
- The frictional force exerted by the table is given as 7 N (opposite to the motion).
- The net force on the block is given by: net force = force due to weight - frictional force.
- Net force = 49 N - 7 N = 42 N.

2. For the 2 kg block:
- The weight of the block is given by: weight = mass * gravity.
- Weight = 2 kg * 9.8 m/s^2 = 19.6 N (downwards).

Since both blocks are attached by a light string passing over a frictionless pulley, their accelerations will be the same. Let's assume the acceleration is a.

3. Using Newton's second law for each block, we can write the following equations:

For the 5 kg block:
- Net force = mass * acceleration.
- 42 N = 5 kg * a.
- a = 42 N / 5 kg = 8.4 m/s^2.

For the 2 kg block:
- Net force = mass * acceleration.
- 19.6 N - tension = 2 kg * a.
- tension = 19.6 N - 2 kg * a = 19.6 N - 2 kg * 8.4 m/s^2 = 19.6 N - 16.8 N = 2.8 N.

Therefore, the tension in the string is 2.8 N.

To calculate the tension in the string, we can use Newton's second law of motion, which states that the sum of all the forces acting on an object is equal to the product of its mass and acceleration.

First, let's calculate the acceleration of the system. Since the mass on the table experiences a frictional force, we need to take that into account.

The frictional force (Ff) can be calculated using the equation Ff = μ * Fn, where μ is the coefficient of friction and Fn is the normal force. The normal force (Fn) is equal to the weight of the block, which is 5 kg * 9.8 m/s^2 (acceleration due to gravity).

Given that the frictional force is 7 N, we can rearrange the equation to solve for the coefficient of friction:

7 N = μ * (5 kg * 9.8 m/s^2)
μ = 7 N / (5 kg * 9.8 m/s^2)
μ ≈ 0.143

Now that we have the coefficient of friction, we can calculate the force (F) acting on the 5 kg mass:

F = μ * Fn
F = 0.143 * (5 kg * 9.8 m/s^2)
F ≈ 7.01 N

Since the force acting on the 5 kg mass is in the opposite direction to the tension in the string, the net force (Fnet) can be calculated as:

Fnet = T - F
where T is the tension in the string.

Using Newton's second law:

Fnet = m * a
where m is the mass and a is the acceleration of the system.

In this case, the mass (m) is the sum of the two masses (5 kg + 2 kg), and the acceleration (a) is the same for both masses.

Fnet = (5 kg + 2 kg) * a

Now, we can rearrange the equation to solve for the tension (T):

T = Fnet + F
T = [(5 kg + 2 kg) * a] + 7.01 N

To get the value of acceleration (a), we can assume that the 2 kg mass is moving downwards. Therefore, it experiences an acceleration due to gravity (9.8 m/s^2). Since the string is light and passes over a frictionless pulley, the 5 kg mass will also have the same acceleration.

Plugging in the values:

T = [(5 kg + 2 kg) * 9.8 m/s^2] + 7.01 N

T ≈ 68.59 N

Therefore, the tension in the string is approximately 68.59 N.