The mass of the larger block

Question

In the diagram below, a worker drags two blocks on a frictionless surface. The work pulls on the first cord, which is connected to the smaller block and at an angle of q = 32.0^O above the horizontal, with a force of 372 N. The blocks accelerate at a rate of 0.628 m/s² and the mass of the smaller block is 130 kg. Find:

  (a) the tension in the cord that connects the two blocks;

  (b) the mass of the larger block. 

Enter your answer to Part (b), in kg, in the box below.

  

Answer 1

I can't download the picture so I'm guessing the diagram looks like this:

Because the force is applied at an angle, only a portion of it helps to accelerate the blocks in the x-direction which I've denoted as F_x.

Using trig, cos(32°) = Fx/372 or Fx = 372cos(32°) = 315.47 N

This is the force that accelerates both blocks and since F = ma, we can say that

315.47 = (130 + m_LB)(0.628)

315.47/0.628 = 130 + m_LB

502.35 = 130 + m_LB

mLB = 502.35 - 130 = 372 kg

Looking at a free-body diagram of the larger block, the only force in the direction of motion is to the right and is the tension in the cord. The larger block is accelerating at the same rate the smaller block is because they move together. Since F = ma, then T = m_LB(0.628) and we just calculated the mass of the larger block so T = 234 N