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Question

Okay so there are two blocks (m1 = 1kg & m2 = .5kg), lying together on a frictionless surface. A constant force of .10N is applied to block m1, pushing both blocks across the table to the right 3.33m in 10 seconds. Then a force of -.10N pushes on block m2, pushing both blocks to the left -3.33 meters in 10 seconds.
What is the acceleration magnitude of the 2 blocks?
In the first scenario, 2 horizontal forces act on m1: the external force and the contact force (P21) acting to the left. Newton's law gives: F - P21 = m1a. The applied force must make both blocks accelerate. Solve the expression for the contact force acting on m1:
N
On block m2, only the contact force (P12) from m1 is acting on it. Thus P12 = m2a. Solve this for the contact force acting on m2: N.
Now reverse the scenario, with the force acting from the left. Solve for the contact force P12, and the contact force acting on m1.

Thanks for the help!

Answer 1

With constant acceleration, we can say x = 0.5*at^2. We know that x = 3.33 m and t = 10 s, so (3.33 m) = 0.5*a(10 s)^2 ==> a = 2*(3.33 m) / (10 s)^2 = 0.0666 m/s^2.

If the contact force of m2 on m1 is P21, then F - P21 = m1a as you wrote. So P21 = F - m1a = (0.10 N) - (1 kg)(0.0666 m/s^2) = 0.0334 N, which is directed towards the left. At the same time, P12 = m2a = (0.5 kg)(0.0666 m/s^2) = 0.0333 N, directed towards the right. These forces are equal an opposite, differing in the fourth decimal place due to rounding in the distance we were given, since the blocks actually would have moved (10/3) m, or 3.333 (repeating).

Repeating in the opposite direction, F - P12 = m2a ==> P12 = F - m2a = (0.10 N) - (0.5 kg)(0.0666 m/s^2) = 0.0667 N to the right and P21 = m1a = (1 kg)(0.0666 m/s^2) = 0.0666 N to the left, again equal and opposite.