TWO BLOCKS ARE CONNECTED BY A LIGHT STRING THAT PASSES OVER A FRICTIONLESS PULLEY AS SHOWN IN THE DIAGRAM. THE COEFFICIENT OF KINETIC FRICTIONBETWEEN THE 6.0KG BLOCK AND THE SURFACE IS 0.25.FIND THE ACCELERATION OF THE 8-KG BLOCK AND THE TENSION OF THE STRING.

To find the acceleration of the 8 kg block and the tension of the string, we need to apply Newton's laws of motion and consider the forces acting on each block.

Let's start by analyzing the forces acting on the blocks:

First, consider the 6.0 kg block:
- The force of gravity acting downward is given by: F_gravity_6kg = m_6kg * g, where m_6kg is the mass of the 6 kg block (6.0 kg) and g is the acceleration due to gravity (9.8 m/s^2).
- The normal force (N) acting upward is equal to the force of gravity, since the block is on a flat surface and there is no vertical acceleration.
- The force of kinetic friction acting in the opposite direction of motion is given by: F_friction = u * N, where u is the coefficient of kinetic friction (0.25) and N is the normal force.

Next, consider the 8 kg block:
- The force of gravity acting downward is given by: F_gravity_8kg = m_8kg * g, where m_8kg is the mass of the 8 kg block (8.0 kg) and g is the acceleration due to gravity (9.8 m/s^2).
- The tension force in the string acts upward.

Since the blocks are connected by a light string passing over a frictionless pulley, the tension in the string will be the same on both sides.

Now, let's use Newton's second law of motion (F_net = m * a) to find the acceleration of the system.

For the 6.0 kg block:
- The net force acting on the 6.0 kg block is given by: F_net_6kg = T - F_friction, where T is the tension in the string.
- Substituting the expressions for F_friction and F_gravity_6kg: F_net_6kg = T - u * N - F_gravity_6kg.

For the 8 kg block:
- The net force acting on the 8 kg block is given by: F_net_8kg = F_gravity_8kg - T.

Since the blocks are connected, they will have the same acceleration (a). Therefore, we can equate the net forces on both blocks:

T - u * N - F_gravity_6kg = F_gravity_8kg - T.

Now, let's plug in the given values and solve the equation:

F_gravity_6kg = (6.0 kg) * (9.8 m/s^2) = 58.8 N
F_gravity_8kg = (8.0 kg) * (9.8 m/s^2) = 78.4 N
N = F_gravity_6kg = 58.8 N
u = 0.25

T - (0.25 * 58.8 N) - 58.8 N = 78.4 N - T.

Now, rearrange the equation to solve for T:

T + T = 78.4 N + 58.8 N + 0.25 * 58.8 N.
2T = 137.2 N.
T = 68.6 N.

Now, plug in the value of T to find the acceleration (a):

68.6 N - (0.25 * 58.8 N) - 58.8 N = (8.0 kg + 6.0 kg) * a.

Simplifying the equation:

68.6 N - (0.25 * 58.8 N) - 58.8 N = 14.0 kg * a,
68.6 N - 14.7 N - 58.8 N = 14.0 kg * a,
-4.9 N = 14.0 kg * a.

Now, solve for a:

a = -4.9 N / 14.0 kg,
a ≈ -0.35 m/s^2.

The acceleration of the 8 kg block is approximately -0.35 m/s^2 (negative sign indicating it's moving opposite to the positive direction) and the tension in the string is approximately 68.6 N.