Ask Question
13 January, 11:10

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. 13 January, 13:40
    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?
Find 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 you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers