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A horizontal force of 300N pulls two block of masses m1=10kg and M2=20 kg which are connected by a lighter inextensible string and lying on the friction less horizontal surface find the acceleration of each other?
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A horizontal force of 300N pulls two block of masses m1=10kg and M2=20...
Acceleration of the Blocks

To find the acceleration of the blocks, we can use Newton's second law of motion, which states that the force acting on an object is equal to the product of its mass and acceleration.

1. Define the Forces
First, let's define the forces acting on the blocks:
- F: the applied force of 300N
- T: the tension in the string connecting the blocks
- m1: mass of the first block (10kg)
- m2: mass of the second block (20kg)
- a1: acceleration of the first block
- a2: acceleration of the second block

2. Equations of Motion
We need to write down the equations of motion for both blocks, taking into account the forces acting on them.

For the first block (m1):
- The applied force F is acting to the right.
- The tension T is acting to the left.

Using Newton's second law, we have:
F - T = m1 * a1 ...(1)

For the second block (m2):
- The tension T is acting to the right.
- There are no other horizontal forces acting on the second block.

Using Newton's second law, we have:
T = m2 * a2 ...(2)

3. Solve the Equations
We have two equations with two unknowns (a1 and a2). We can solve these equations simultaneously to find the values of acceleration for each block.

From equation (2), we can express the tension T in terms of a2:
T = m2 * a2

Substituting this value of T in equation (1), we get:
F - m2 * a2 = m1 * a1

Rearranging the equation, we have:
a1 = (F - m2 * a2) / m1

Now, we substitute the given values: F = 300N, m1 = 10kg, and m2 = 20kg.

a1 = (300N - 20kg * a2) / 10kg

4. Solve for a2
To solve for a2, we need an additional equation. In this case, we can consider the force balance on the second block.

Since there are no horizontal forces acting on the second block, the tension T is equal to the force exerted by the first block on the second block.

T = m2 * a2

Substituting the given values, we have:
m2 * a2 = T = m2 * a1

Dividing both sides by m2, we get:
a2 = a1

5. Substitute for a2
Now, we can substitute the value of a2 in terms of a1 into the equation for a1 that we derived earlier.

a1 = (300N - 20kg * a1) / 10kg

Multiplying both sides by 10kg, we have:
10kg * a1 = 300N - 20kg * a1

Combining like terms, we get:
30kg * a1 = 300N

Dividing both sides by 30kg, we find:
a1 = 10N/kg

Since a1 = a2, the acceleration of both blocks is 10N/kg or 10m/s².
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A horizontal force of 300N pulls two block of masses m1=10kg and M2=20 kg which are connected by a lighter inextensible string and lying on the friction less horizontal surface find the acceleration of each other?
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