A 42 kg child on a swing with a radius of 3.50 m. At the lowest point, the seat exerts an upward force of 550 N on the child. What is the speed of the child at this point? 

Answer given in explanation. 

Step-by-step explanation

According to force diagram force acting are tension ( T ) and mass of body ( mg ) . The child's Path is an arc of Circle, At the lowest point the child's motion is changing in direction . He moves with centripetal acceleration even as hig Speed is not changing and his tangential acceleration is zero. According to Newton's Second law. T- mg = ma Given T = 550 N Y = 3.5 m. m = 42kg T - mg = mv2 550 - 42 x9.8 , = 42 x 12 3.5 138. 4 = 12 2 V = 138.4 12 V = 3. 396 m/sec.

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