Motion on a Rough Inclined Plane · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Splitting weight on a slope

Mathematics · Mechanics · ages 17-18
Name ______________________   Date ____________

  1. A block weighs 50 N on a slope with sin = 0.6. What is the component of its weight pulling it down the slope, in N?

    Answer: ______________

  2. A block weighs 50 N on a slope with sin equal to 0.6. What is the component of its weight pulling it down the slope, in N?

    Answer: ______________

  3. A 5 kg block rests on a slope with cos equal to 0.8. Take g equal to 10 metres per second squared. What is the normal reaction?

    • 40 N
    • 30 N
    • 50 N

  4. A block slides down a rough slope. Which way does friction act?

    • Down the slope, with the motion
    • Up the slope, against the motion
    • Out of the slope at right angles

  5. A block is on the point of slipping on a slope tilted at 45 degrees. Since mu equals tan of the slip angle, what is mu?

    Answer: ______________

  6. For a body on the point of slipping, how is mu linked to the angle of friction?

    • mu equals sin of the angle
    • mu equals cos of the angle
    • mu equals tan of the angle

  7. A 2 kg block slides down a slope with sin 0.6 and cos 0.8. Friction coefficient mu is 0.25. Take g equal to 10 metres per second squared. What is its acceleration in metres per second squared?

    Answer: ______________

  8. A 2 kg block slides down a slope with sin = 0.6 and cos = 0.8. Friction coefficient mu = 0.25. Take g = 10 m/s2. What is its acceleration in m/s2?

    Answer: ______________

  9. A slope has sin 0.6 and cos 0.8. A block just slips there. Since mu equals tan, give mu as opposite over adjacent.

    Answer: ______________

  10. For a body on the point of slipping, how is mu linked to the angle of friction?

    • mu = tan of the angle
    • mu = sin of the angle
    • mu = cos of the angle
    • mu = the angle in degrees
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Answer key

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Splitting weight on a slope W1-mt_0rBy66HFNO-s1

  1. 30 · Down slope component is weight times sin: 50 by 0.6 = 30.
  2. 30 · The down slope component is weight times sin: 50 times 0.6.
  3. 40 N · Across the slope, normal equals mg cos: 5 times 10 times 0.8.
  4. Up the slope, against the motion · Friction opposes the actual motion, so it points up while the block slides down.
  5. 1 · tan 45 = 1, so the angle of friction gives mu = 1.
  6. mu equals tan of the angle · Balancing along and across the slope at the limit gives mu equals tan.
  7. 4 · Net pull per mass is 10 times (0.6 minus 0.25 times 0.8), which is 4.
  8. 4 · Net pull is 10 by (0.6 - 0.25 by 0.8) = 4.
  9. 0.75 · tan is sin over cos, and 0.6 divided by 0.8 is 0.75.
  10. mu = tan of the angle · Balancing along and across the slope at the slip point gives mu = tan.
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