Two blocks connected by a cord passing over a small, frictionless pulley rest on frictionless planes as shown Find the acceleration of the blocks and the tension in the cord

Answer 1

Let's denote the mass of the block on the left as m1 and the mass of the block on the right as m2. The acceleration of both blocks will be the same, let's call it a.

Since the system is frictionless, the only forces acting on the blocks are the tension in the cord and the force of gravity. Let's consider the block on the left first.

For block m1:
- Tension force in the cord is pulling it to the right
- Force of gravity is pulling it down

The net force acting on block m1 is given by:
m1*g - T = m1*a

For block m2:
- Tension force in the cord is pulling it to the left
- Force of gravity is pulling it down

The net force acting on block m1 is given by:
T = m2*a

Since the blocks are connected by a cord, the tension in the cord will be the same for both blocks.

Now, let's combine the equations and solve for the acceleration a:
m1*g - T = m1*a
T = m2*a

First, we substitute the second equation into the first one:
m1*g - m2*a = m1*a

Solving for a:
a = m1*g / (m1 + m2)

Now, we can plug in the values of m1 and m2, along with the acceleration due to gravity g, to find the acceleration of the blocks.

Once we have the acceleration, we can also calculate the tension in the cord by using the second equation:
T = m2*a

I hope this helps! Let me know if you have any further questions.