Ask Question
14 January, 04:19

When the 2.8-kg bob is given a horizontal speed of 1.5 m/s, it begins to rotate around the horizontal circular path A. The force F on the cord is increased, the bob rises and then rotates around the horizontal circular path B. a) Determine the speed of the bob around path B

b) Find the work done by force F.

+5
Answers (1)
  1. 14 January, 06:01
    0
    The speed is the same at 1.5 m/s while

    The work done by the force F is 0.4335 J

    Explanation:

    Here we have angular acceleration α = v²/r

    Force = ma = 2.8 * 1.5²/r₁

    and ω₁ = v₁/r₁ = ω₂ = v₁/r₂

    The distance moved by the force = 600 - 300 = 300 mm = 0.3 m

    If the velocity is constant

    The speed is 1.5 m/s while the work done is

    2.8 * 1.5²1 / (effective radius) * 0.3

    r₁ = effective radius

    2.8*9.81 = 2.8 * 1.5²/r₁

    r₁ = 0.229

    The work done by the force = 2.8 * 1.5²*1/r₁ * 0.3 = 0.4335 J
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “When the 2.8-kg bob is given a horizontal speed of 1.5 m/s, it begins to rotate around the horizontal circular path A. The force F on the ...” in 📙 Engineering 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