You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo moves down a distance 0.32 m. The mass of the yo-yo is 0.062 kg, and it was initially moving downward with speed 2.9 m/s.(a) What is the increase in the translational kinetic energy of the yo-yo?(b) What is the new speed of the yo-yo?(c) What is the increase in the rotational kinetic energy of the yo-yo?

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

hence, the change in Er is about 1.52J times the initial rotational energy

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