Question

A bowling ball weighing 71.2 N (=mg) is attached to the ceiling by a 3.40 m rope. The ball is pulled to one side and released; it then swings back and forth as a pendulum. The ball is initially pulled back so it is a vertical height of 2.2 m above the lowest point of the ball's swing.

1. When the ball is at its lowest point in its swing, what is the acceleration of the bowling ball, in magnitude and direction? Enter a positive value if the acceleration is upward and negative value if the acceleration is downward. (m/s^2)

2. At this instant, what is the tension in the rope? Express your answer with the appropriate SI units. (N)

Answer 1

uring conservation of energy Ei=E2 ngh= mv2 2 Th v² = 29h = (2x9.842.2/(m/s2 V?= 43.12 Cm/s)? ung thu H) acs centrepetal acceleration = 4* = (43.2 = + 12.68 m/s2 tangentiel aceleration comis? a = Net acceleration - Janatat? =((2.68)²+02 m/s² = +12-68 m/s2 1 / 63 m/s upward at= AT (h) T= tension in rope=img + mult = m (4+) 'mv² 71-2 (9+4+) 9.8+1268)N 9.8 g T = 163.348