A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangentialacceleration. What is the magnitude of this acceleration if the kinetic energy of theparticle becomes equal to 32 × 10−4 J by the end of the third revolution after thebeginning of the motion?
Question
A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangentialacceleration. What is the magnitude of this acceleration if the kinetic energy of theparticle becomes equal to 32 × 10−4 J by the end of the third revolution after thebeginning of the motion?
Solution
To solve this problem, we need to use the formula for kinetic energy, which is KE = 1/2 * m * v^2, where m is the mass and v is the velocity. We also need to use the formula for acceleration, which is a = v^2 / r, where r is the radius.
Step 1: Convert the given values to SI units. The mass m = 10 g = 0.01 kg, the radius r = 6.4 cm = 0.064 m, and the kinetic energy KE = 32 × 10^-4 J = 0.0032 J.
Step 2: Substitute the given values into the kinetic energy formula and solve for v. We get 0.0032 = 1/2 * 0.01 * v^2, which simplifies to v^2 = 0.0032 / 0.005 = 0.64. Therefore, v = sqrt(0.64) = 0.8 m/s.
Step 3: Substitute the values of v and r into the acceleration formula to find the acceleration. We get a = 0.8^2 / 0.064 = 0.64 / 0.064 = 10 m/s^2.
Therefore, the magnitude of the acceleration of the particle is 10 m/s^2.
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