Question

The same force that gives the standard 1 kg mass an acceleration of 1.00 m/s2 acts first on body A, producing an acceleration of 0.530 m/s2, and then on body B, producing an acceleration of 0.344 m/s2. Find the acceleration produced when A and B are attached and the same force is applied.

(The answer given is 0.209 m/s^2; HOW IS THIS CALCULATED... steps please). I tried and dont understand).

Answer 1

To find the combined acceleration when A and B are attached, we calculate the individual forces acting on A and B using Newton's second law, F = ma. Then, we express the total force as the sum of the individual forces and calculate the combined acceleration using the formula a = (FA + FB) / (mA + mB). Plugging in the given values, we find that the combined acceleration is 0.209 m/s2.

Step-by-step explanation:

To find the combined acceleration when A and B are attached and the same force is applied, we need to consider the individual accelerations of A and B. First, let's find the individual forces acting on A and B using Newton's second law, F = ma. For body A, the force FA is given by FA = mA*aA, where mA is the mass of body A and aA is its acceleration. Similarly, for body B, the force FB is given by FB = mB*aB.

Now, when A and B are attached, they experience the same force F. This force can be split into two components, one acting on body A and the other acting on body B. We can write this as F = FA + FB. Rearranging this equation, we have a = (FA + FB) / (mA + mB), where a is the acceleration when A and B are attached.

Substituting the values given in the question, we have a = (mA*aA + mB*aB) / (mA + mB). Plugging in the values, we get a = (1 kg * 0.530 m/s2 + 1 kg * 0.344 m/s2) / (1 kg + 1 kg) = 0.209 m/s2.

Answer 2

The acceleration produced is
`0.209 m/s^(2)`

Step-by-step explanation:

By the second law of Newton, the force F is equal to:

F = ma

Where m is the mass of the object and a is the acceleration produced. So if The force gives the standard 1 kg mass an acceleration of 1.00 m/s2, that means that the force apply on A and B is equal to:

F = (1Kg) * (1.00m/s2) = 1 N

Then, if this force on A produce an acceleration of 0.530
`m/s^(2)`, the mass of A is:

`F = m_A*a_A \\1 N = m_A* (0.530 m/s^(2) )\\(1N)/(0.530m/s^(2) ) =m_A\\\\1.887Kg = m_A`

At the same way, if this force on B produce an acceleration of 0.344
`m/s^(2)`, the mass of B is:

`F = m_B*a_B \\1 N = m_B* (0.344 m/s^(2) )\\(1N)/(0.344m/s^(2) ) =m_B\\\\2.907Kg = m_B`

Therefore, if they are attached and the same force is applied, the acceleration is:

`F=(m_A+m_B)*a\\1N=(1.887Kg + 2.907Kg)*a\\1N = 4.794 Kg *a\\(1N)/(4.794Kg)=a\\ 0.209 m/s^(2) =a`