Two blocks with masses m1 = 1.40 kg and m2 = 3.00 kg are connected by a massless string, as shown in the Figure. They are released from rest. The coefficent of kinetic friction between the upper block and the surface is 0.390.In the figure m1 is on top of a horizontal table with a pully connecting m1 and m2 which is hanging off the edge of the table.



Assume that the pulley has a negligible mass and is frictionless, and calculate the speed of the blocks after they have moved a distance 20.0 cm.

To find the speed of the blocks after they have moved a distance of 20.0 cm, we can use the principles of Newton's laws and the concept of work and energy.

First, let's calculate the force of friction acting on the upper block. We can use the equation:

Frictional force = coefficient of kinetic friction * Normal force

The normal force is equal to the weight of the upper block, which is given by:

Normal force = mass of upper block * acceleration due to gravity

Normal force = m1 * g

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Now, we can calculate the frictional force:

Frictional force = (coefficient of kinetic friction) * (m1 * g)

Next, we need to find the net force acting on the system. Since the pulley is frictionless and has negligible mass, the tension in the string is the same throughout the length of the string. Thus, the tension is equal to the force of friction on the upper block.

Now, the net force on the upper block is given by:

Net force = applied force - frictional force

Since the upper block is released from rest, there is no applied force, so the net force reduces to:

Net force = - frictional force

Now, we can use Newton's second law to find the acceleration of the system:

Net force = mass of upper block * acceleration

-frictional force = m1 * a

Solving for acceleration (a):

a = (-frictional force) / m1

Now, the work done on the system can be calculated by:

Work = force * distance

Since the force acting on the system is the tension in the string and the distance is 20.0 cm (or 0.20 m), the work done is:

Work = tension * distance

Finally, we can use the work-energy principle to calculate the change in kinetic energy of the system:

Change in kinetic energy = work done

Since the system starts from rest, the initial kinetic energy is zero, and the final kinetic energy is equal to the kinetic energy of the two blocks after they have moved a distance of 20.0 cm.

Change in kinetic energy = kinetic energy of the blocks

The kinetic energy of an object is given by:

Kinetic energy = (1/2) * mass * velocity^2

Substituting the masses of the blocks, we have:

Change in kinetic energy = (1/2) * (m1 + m2) * velocity^2

Now, equating the work done to the change in kinetic energy, we have:

Tension * distance = (1/2) * (m1 + m2) * velocity^2

Solving for velocity, we get:

velocity = sqrt((2 * Tension * distance) / (m1 + m2))

Substituting the values given in the problem:

velocity = sqrt((2 * frictional force * distance) / (m1 + m2))

Now, you can substitute the values of the variables given in the problem (m1 = 1.40 kg, m2 = 3.00 kg, coefficient of kinetic friction = 0.390, and distance = 20.0 cm) into the equation to find the speed of the blocks after they have moved a distance of 20.0 cm.