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KIM [24]
3 years ago
12

The 1.0-kg collar slides freely on the fixed circular rod. Calculate the velocity v of the collar as it hits the stop at B if it

is elevated from rest at A by the action of the constant 58-N force in the cord. The cord is guided by the small fixed pulleys.
Physics
1 answer:
soldi70 [24.7K]3 years ago
7 0

Answer:

6.21 m/s

Explanation:

Using work energy equation then

U_{1-2}=T_B- T_A\\58d-mgh=0.5m(v_b^{2}-v_a^{2})

where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity

Substituting 1 Kg for m, 0.4m for h, v_a as 0, 9.81 for g then

58(\sqrt{0.4^{2}+0.3^{2}}-0.1)-(1\times 9.81\times 0.4)=0.5\times 1\times (v_b^{2}-v_a^{2})\\19.276=0.5\times 1v_b^{2}\\v_b=6.209025688 m/s\approx 6.21 m/s

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