A .001 kg bead slides without friction around a loop-the-loop. The bead is released from a height of 20.6 m from the bottom of t
1 answer:
Answer:
11.37 m/s
Explanation:
We are given;
Mass; m = 0.001 kg
Initial velocity; v1 = 0 m/s (since it was released from rest)
Initial height; h1 = 20.6m
From the attached image of the bead in the loop, we can see that height at A; h2 = twice the radius = 2R = 2 × 7 = 14 m
g = 9.8
To solve for the speed at point, we will use formula for conservation of energy. Thus;
½m(v1)² + mgh1 = ½m(v2)² + mgh2
Divide through by m to get;
½(v1)² + gh1 = ½(v2)² + gh2
Plugging in the relevant values;
½(0²) + (9.8 × 20.6) = ½(v2)² + (9.8 × 14)
Rearranging, we have;
½(v2)² = ½(0²) + (9.8 × 20.6) - (9.8 × 14)
Simplifying, we have;
½(v2)² = 64.68
Multiply both sides by 2 to get;
(v2)² = 2 × 64.68
(v2)² = 129.36
v2 = √129.36
v2 = 11.37 m/s
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