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Where a force of 200N pulls two blocks together(as one system) across a horizontal table top(µ=0.800)

$m_A$ = 5.00kg, $m_B$ = 10.0kg

  1. Find the acceleration of the system.
  2. Find f$_k$ between B and A

I found a to be 5.485 m/s² which agrees with the textbook's 5.5m/s².

The textbook says b is 173N but I can't seem to get a number even close to that. How does one go about solving this kind of equation, please provide calculations or formulas in the order they are needed, insteed of just abstract steps.

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2 Answers

up vote 1 down vote accepted

If both block A, and B are moving together as a system, the two blocks will not have a kinetic friction between the two of them (because they are stationary to each other). Draw your free-body diagram of both blocks individually, and write an expression for all the forces acting on the each block. Share what you find by editing your question, so that we may know why you might not be getting the answer you desire.

$\sum F = ma$

Edit: You did not explain the question well enough, but I can see that the friction is indeed equal to 172.5 N assuming that mass B is on top of mass A, and the force is applied on the top mass.

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Ok I figured it out, sorry the fact that block A is on top of block B is important.

F$_a$ = 5.485m/s²(5kg) = 27.425N

$\Sigma F_x = 200N ∴ f_s= 200N - F_a = 200N -27.435 =172.575$

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