# Two blocks with masses m1 = 1.20 kg and m2 = 2.90 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 0.300. Assume that the pulley has a negligible mass and is frictionless, and calculate the speed of the blocks after they have moved a distance 74.0 cm.

## The acceleration of the blocks is given by:

a = (Fnet)/(m1 + m2)

Fnet = Fg1 + Fg2 - Ff

Fg1 = m1g

Fg2 = m2g

Ff = μk(m1 + m2)g

where μk is the coefficient of kinetic friction.

Substituting the given values, we get:

a = (m1g + m2g - μk(m1 + m2)g)/(m1 + m2)

a = (1.20 kg * 9.8 m/s2 + 2.90 kg * 9.8 m/s2 - 0.300 * (1.20 kg + 2.90 kg) * 9.8 m/s2)/(1.20 kg + 2.90 kg)

a = 0.837 m/s2

The speed of the blocks after they have moved a distance of 74.0 cm is given by:

v = √(2as)

v = √(2 * 0.837 m/s2 * 0.740 m)

v = 1.45 m/s

## To find the speed of the blocks after they have moved a distance of 74.0 cm, we can break this problem down into several steps.

Step 1: Calculate the acceleration of the system.

Using Newton's second law, we can write the following equations of motion:

For block 1:

m1*a = m1*g - T1 - f1

For block 2:

m2*a = T2 - m2*g - f2

Where:

m1 = mass of block 1 = 1.20 kg

m2 = mass of block 2 = 2.90 kg

a = acceleration of the system

g = acceleration due to gravity = 9.8 m/s^2

T1 = tension in the string connected to block 1

T2 = tension in the string connected to block 2

f1 = frictional force between block 1 and the surface

f2 = frictional force between block 2 and the surface

The tensions T1 and T2 are equal because the string connecting the blocks is massless. So we can simplify the equations as follows:

m1*a = m1*g - T - f1 (Equation 1)

m2*a = T - m2*g - f2 (Equation 2)

Step 2: Find the frictional forces.

The frictional forces can be calculated using the equation:

frictional force = coefficient of friction * normal force

The normal force on each block can be calculated considering that there is no vertical acceleration:

For block 1:

normal force 1 = m1 * g

For block 2:

normal force 2 = m2 * g

So the frictional forces are:

f1 = coefficient of kinetic friction * normal force 1 (Equation 3)

f2 = coefficient of kinetic friction * normal force 2 (Equation 4)

Step 3: Substitute the frictional forces back into the equations of motion.

Substituting equations 3 and 4 back into equations 1 and 2, we get:

m1*a = m1*g - T - (coefficient of kinetic friction * normal force 1) (Equation 5)

m2*a = T - m2*g - (coefficient of kinetic friction * normal force 2) (Equation 6)

Step 4: Solve the system of equations.

Solving equations 5 and 6 simultaneously, we can find the acceleration of the system.

Step 5: Calculate the speed of the blocks.

The final speed v can be found using the equation:

v^2 = u^2 + 2*a*s

Where:

u = initial speed = 0 (as the blocks are released from rest)

a = acceleration of the system (from step 4)

s = distance traveled by the blocks = 74.0 cm = 0.74 m

Solving this equation will give us the value of v, which is the speed of the blocks after they have moved a distance of 74.0 cm.

## To solve this problem, we can use the principles of mechanics and apply Newton's laws of motion.

1. First, let's determine the net force acting on each block:

- For the upper block with mass m1 = 1.20 kg, the net force is the force of tension in the string (T1) minus the force of friction (Ff1).

- For the lower block with mass m2 = 2.90 kg, the net force is the force of tension in the string (T2) plus the force of friction (Ff2).

2. The force of tension in the string (T1) and (T2) is the same because the string is inextensible and has negligible mass. We'll call it T.

3. The force of friction (Ff) can be calculated using the formula Ff = μk * N, where μk is the coefficient of kinetic friction and N is the normal force. The normal force is the force exerted by the surface perpendicular to the direction of motion.

4. The normal force (N) acting on each block can be calculated using the formula N = mg, where g is the acceleration due to gravity (approximately 9.8 m/s²).

5. Now, we can write the equations of motion for each block. Since the force of tension (T1) and (T2) have opposite directions, we'll consider them positive for one block and negative for the other.

- For the upper block:

Net force = T1 - Ff1 = m1 * a1

T - μk * N1 = m1 * a1

- For the lower block:

Net force = T2 + Ff2 = m2 * a2

T + μk * N2 = m2 * a2

6. Note that the acceleration (a1 = a2 = a) is the same for both blocks because they are connected by a massless string.

7. The distance covered by the blocks is given as 74.0 cm. To relate distance and acceleration, we can use the equation of motion: s = ut + 0.5 * a * t², where s is the distance, u is the initial velocity (0 m/s in this case), and t is the time taken to cover the distance.

8. Rearranging the equation, we get t = √(2s / a).

9. Substituting back into the equations of motion, we have two equations for the unknowns T and a:

For the upper block: T - μk * N1 = m1 * √(2s / a)

For the lower block: T + μk * N2 = m2 * √(2s / a)

10. Solving these equations simultaneously will give us the values for T and a.

11. Finally, we can calculate the speed of the blocks using the equation v = u + at, where v is the final velocity, u is the initial velocity (0 m/s), and t is the time taken to cover the distance.

By following these steps, we can find the speed of the blocks after they have moved a distance of 74.0 cm.