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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.

Respuesta :

Answer:

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

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