Block B is on a resistance-free surface and a horizontal force of 4.54 N acts on the block. This force produces an acceleration of 2.44 m/s/s. Block A, which has a mass of 4.38 kg, is then dropped onto Block B as shown in the "after" picture. The same force continues to act. What is the acceleration of the combination of blocks? (Assume that the second block does not slide on the first block.)
if you (about) double the mass
you will (about) halve the acceleration
a = f / m
There is a hint: First, find the mass of the original block. Then use Newton's second law equation to determine the acceleration of the combination of two masses.
mass of block B = F/a = 4.54 / 2.44 =1.86 kg
new total mass = 1.86 + 4.38 = 6.24 kg
a = F/m = (4.54 / 6.24) m/s^2
To find the acceleration of the combination of blocks, we can use Newton's second law of motion. This law states that the force applied to an object is equal to the mass of the object multiplied by its acceleration.
First, let's find the mass and acceleration of Block B.
Given:
Force on Block B = 4.54 N
Acceleration of Block B = 2.44 m/s²
We know that mass (m) is related to force (F) and acceleration (a) by the equation F = ma. Rearranging this equation gives us m = F/a.
Substituting the given values, we have:
m(B) = 4.54 N / 2.44 m/s² = 1.86 kg
Block B has a mass of 1.86 kg.
Now, let's focus on Block A.
Given:
Mass of Block A = 4.38 kg
Block A is dropped onto Block B, so it essentially adds to the mass of the system.
The total mass of the combination of blocks is:
m(total) = m(A) + m(B) = 4.38 kg + 1.86 kg = 6.24 kg
Therefore, the total mass of the combination of blocks is 6.24 kg.
Since the same force continues to act on the combination of blocks, we can use Newton's second law again to find the acceleration.
Using the equation F = ma, we have:
4.54 N = (m(total)) * a
Substituting the total mass of 6.24 kg, we can solve for the acceleration:
a = 4.54 N / 6.24 kg ≈ 0.728 m/s²
Therefore, the acceleration of the combination of blocks is approximately 0.728 m/s².