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

Answer:

The acceleration produced is $0.209 m/s^{2}$

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:

$F = m_A*a_A \\1 N = m_A* (0.530 m/s^{2} )\\\frac{1N}{0.530m/s^{2} } =m_A\\\\1.887Kg = m_A$

At the same way,  if this force on B produce an acceleration of 0.344 $m/s^{2}$, the mass of B is:

$F = m_B*a_B \\1 N = m_B* (0.344 m/s^{2} )\\\frac{1N}{0.344m/s^{2} } =m_B\\\\2.907Kg = m_B$

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

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