A ball with a mass of 200 g is tied to a light string that has a length of 2.80

Question

A ball with a mass of 200 g is tied to a light string that has a length of 2.80 m. The end of the string is tied to a hook, and the ball hangs motionless below the hook. Keeping the string taut, you move the ball back and up until the string makes an angle of 27.0° with the vertical. You then release the ball from rest, and it oscillates back and forth, pendulum style. Use g = 9.80 m/s2.

(a) If we neglect air resistance, what is the highest speed the ball achieves in its subsequent motion? (m/s)

(b) Resistive forces eventually bring the system to rest. Between the time you release the ball and the time the ball comes to a permanent stop, how much work do the resistive forces do? (Use the appropriate sign.) (J)

Explanation / Answer

Mass of ball m = 200 g = 0.2 kg Length of the string l = 2.80 m Angle made by string with vertical = 27.0° (a) From law of conservation of energy               mgh = 1/2 mv2 here height h = l ( 1 - cos )                       = (2.80 m) ( 1 - cos 27.0°)                       = 0.305 m then                gh = (1/2) v2 therefore the maximum speed                v = 2gh                     = 2(9.8 m/s2) (0.305 m)                     = 2.44 m/s ------------------------------------------------------------------------------- (b) From work energy theroem                  W = K                       = 1/2 mv2                       = 1/2 (0.2 kg) (2.44 m/s)2                       = 0.595 J Therefore wokdone by frictional force is W = - 0.6 J                       = 0.595 J Therefore wokdone by frictional force is W = - 0.6 J

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