Final answer:
The radius of the circle in which an 8 kg object moves with constant speed, given a kinetic energy of 35 J and string tension of 322 N, is calculated to be approximately 0.217 meters.
Step-by-step explanation:
The question involves finding the radius of the circle in which an 8 kg object moves with constant speed on a frictionless surface, given its kinetic energy and the tension in the string. To solve this problem, we can use the formulas from uniform circular motion and work-energy principles. The kinetic energy (KE) of the object is given by KE = 1/2 mv², where m is the mass and v is the tangential velocity of the object. Since we know the kinetic energy (35 J), we can solve for v.
Once we have the velocity, we can use the centripetal force formula Fc = mv²/r, where Fc is the centripetal force, which in this case is provided by the tension in the string (322 N), to solve for the radius r of the circle.
We can set the two equations equal to each other and solve for the radius:
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- Calculate velocity: KE = 1/2 mv² → v² = 2KE/m → v = √(2×35/8) → v = √(70/8) → v = √8.75 m/s.
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- Calculate radius: Fc = mv²/r → r = mv²/Fc → r = (8×8.75)/322 → r = 70/322 → r = 0.21739 meters.
Therefore, the radius of the circle is approximately 0.217 meters.