Question

Date . A 200 g ball at the end of a 1.25 m string is twirled in a horizontal circle at constant speed. The ball makes a complete revolution every 0.600 s. (a) Ans: (b) What is the acceleration of the ball? Ans: (c) what is the tension in the string? Ans: What is the speed of the ball? A force of (5.0N, tx) acts on a 6.00 kg object moving along a horizontal surface. The accelenation of the 2. object is (1.50 m/s, +x). (a) What is the magnitude of the resultant force acting? A. 4N B. 6 N (b) What is themagnitude of the friction force? A. 4N 6N C. 9N D. 15N E. 10N F. 24N

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Answer #1

1. w = 2*pi/ T = 2*3.14/0.6 = 10.47 rad/s

a) v = rw = 1.25* 10.47 = 13.08 m/s

b) a = r w^2 = 1.25* 10.47^2 = 137.03 m/s^2

c) tension = ma = 0.2* 137.03 = 27.4 N

==========

2.

a) resultant force acting = 6*1.5 = 9 N

b) magnitude of force = 6 ( 15/6 - 1.5) = 6 N

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Comment in case any doubt.. Goodluck

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