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Tension Extra Drill Problems for AP Physics B & C

based on 3 ratings
By — McGraw-Hill Professional
Updated on Feb 14, 2011

Practice problems and tests cannot possibly cover every situation that you may be asked to understand in physics.  However, some categories of topics come up again and again, so much so that they might be worth some extra review.  And that's exactly what these lessons are for - to give you a focused, intensive review of a few of the most essential physics topics.

Extra drill on difficult but frequently tested topics are:

We call them "drills" for a reason.  They are designed to be skill-building exercises and as such, they stress repetition and technique.  Working through these exercises might remind you of playing scales if you're a musician or of running laps around the field if you're an athlete.  Not much fun, maybe a little tedious, but very helpufl in the long run.

The questions in each drill are all solved essentially the same way.  Don't just do one problem after the other.....rather, do a couple, check to see that your answers are right, and then half an hour or a few days later, do a few more, just to remind yourself of the techniques involved.

Below are tension problems.

How to Do It

Use the following steps to solve these kinds of problems: 1) Draw a free-body diagram for each block; 2) resolve vectors into their components; 3) write Newton's second law for each block, being careful to stick to your choice of positive direction; and 4) solve the simultaneous equations for whatever the problem asks for.

The Drill

In the diagrams below, assume all pulleys and ropes are massless, and use the following variable definitions.

      F = 10 N
      M = 1.0 kg
      μ = 0.2

Find the tension in each rope and the acceleration of the set of masses.

(For a greater challenge, solve in terms of F, M, and μ instead of plugging in values.)

  1. Frictionless
  2. Frictionless

  3. Frictionless
  4. Frictionless

  5. Frictionless
  6. Frictionless

  7. Coefficient of Friction μ
  8. Coefficient of Friction μ

  9.  
  10. Coefficient of Friction μ

  11.  
  12. Coefficient of Friction μ

  13. Frictionless
  14. Frictionless

  15. Frictionless
  16. Frictionless

  17. Frictionless
  18. Frictionless

  19. Coefficient of Friction μ
  20. Coefficient of Friction μ

  21. Coefficient of Friction μ
  22. Coefficient of Friction μ

  23. Frictionless
  24. Frictionless

  25. Frictionless
  26. Frictionless

  27. Coefficient of Friction μ
  28. Coefficient of Friction μ

The Answers (Step-by-Step Solutions to #2 and #5 are below.)

  1. a = 10 m/s2
  2. a = 3.3 m/s2
  3. T = 3.3 N

  4. a = 1.7 m/s2
  5. T1 = 1.7 N

    T2 = 5.1 N

  6. a = 1.3 m/s2
  7. T = 3.3 N

  8. a = 3.3 m/s2
  9. T = 13 N

  10. a = 8.6 m/s2
  11. T1 = 19 N

    T2 = 9 N

  12. a = 3.3 m/s2
  13. T = 6.6 N

  14. a = 6.7 m/s2
  15. T1 = 13 N

    T2 = 10 N

  16. a = 1.7 m/s2
  17. T1 = 5.1 N

    T2 = 8.3 N

  18. a = 6.0 m/s2
  19. T = 8.0 N

  20. a = 8.0 m/s2
  21. T1 = 10 N

    T2 = 4.0 N

  22. a = 5.0 m/s2
  23. T = 15 N

  24. a = 3.3 m/s2
  25. T1 = 13 N

    T2 = 20 N

  26. a = 0.22 m/s2
  27. T1 = 20 N

    T2 = 29 N

Step-by-Step Solution to #2:

Step 1: Free-body diagrams:

Free-body diagram

No components are necessary, so on to Step 3: write Newton's second law for each block, calling the rightward direction positive:

    T – 0 = ma.
    FT = (2m)>a.

Step 4: Solve algebraically. It's easiest to add these equations together, because the tensions cancel:

    F = (3m)a, so a = F/3m = (10 N)/3(1 kg) = 3.3 m/s2.

To get the tension, just plug back into T – 0 = ma to find T = F/3 = 3.3 N.

Step-by-Step Solution to #5:

Step 1: Free-body diagrams:

Tension

No components are necessary, so on to Step 3: write Newton's second law for each block, calling clockwise rotation of the pulley positive:

      (2m)gT = (2m)a.
      Tmg = ma.

Step 4: Solve algebraically. It's easiest to add these equations together, because the tensions cancel:

    mg = (3m)a, so a = g/3 = 3.3 m/s2.

To get the tension, just plug back into Tmg = ma : T = m (a + g) = (4/3)mg = 13 N.

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