Blocks of mass 4, 8, and 24 kg are lined up from left to right in that order on a frictionless surface so each block is touching the next one. A rightward-pointing force of magnitude 15 N is applied to the left-most block.
Suppose now that the left-right order of the blocks is reversed. Now find the magnitude of the force that the leftmost block exerts on the middle one?
To find the magnitude of the force that the leftmost block exerts on the middle one, we can use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.
In this case, when the leftmost block exerts a force on the middle one, the middle block exerts an equal and opposite force on the leftmost block.
Since the leftmost block is now on the right side, we need to take into account the change in direction. The force exerted by the leftmost block on the middle one will be in the opposite direction to its previous force, but its magnitude will remain the same.
Now, let's calculate the magnitude of the force that the leftmost block exerts on the middle one.
Step 1: Calculate the acceleration of the blocks
In this scenario, the force applied to the leftmost block is 15 N. To find the acceleration, we can use Newton's second law of motion:
F = m * a
Where:
F is the force applied,
m is the mass of the block,
a is the acceleration.
For the leftmost block (mass = 4 kg):
15 N = 4 kg * a
a = 15 N / 4 kg
a = 3.75 m/s^2
Step 2: Calculate the force exerted by the leftmost block on the middle one
Now that we know the acceleration, we can calculate the force exerted by the leftmost block on the middle one. The force exerted will be equal to the mass of the middle block multiplied by the acceleration (F = m * a):
For the middle block (mass = 8 kg):
F = 8 kg * 3.75 m/s^2
F = 30 N
Therefore, the magnitude of the force that the leftmost block exerts on the middle one is 30 N.