Ask Question
4 May, 05:03

How much work must you do to push a 12 kg block of steel across a steel table at a steady speed of 1.3 m/s for 7.6 s? the coefficient of kinetic friction for steel on steel is 0.60?

+5
Answers (1)
  1. 4 May, 07:01
    0
    Work is force times distance.

    The distance is 1.3 m/s x 7.6 s = 9.88 m

    the force is only sufficient force to overcome friction. Assuming the table is a level table, the force to overcome friction is µ x normal force = 0.6 x (12 kg) x 9.8 m/s^2 = 70.56 N

    So the work is 70.56 N x 9.88 m = 697.13 J

    The power is simply the work / time = 697.13 J / 7.6 s = 91.7 or 92 Watts
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “How much work must you do to push a 12 kg block of steel across a steel table at a steady speed of 1.3 m/s for 7.6 s? the coefficient of ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers