A pendulum bob is released from some initial height such that the speed of the bob at the bottom of the swing is 3.6 m/s. the acceleration of gravity is 9.81 m/s 2 . what is the initial height of the bob? answer in units of m.

Respuesta :

0.66 m
   Ignoring friction, the sum of the kinetic energy and the potential gravitational energy of the pendulum bob remains constant. At the bottom of the swing, the bob has 0 potential gravitational energy and maximum kinetic energy. At the top of the swing, the situation is reversed with the bob having 0 kinetic energy. So at the initial height of the bob, it's gravitational potential energy has to equal the kinetic energy at the bottom of the swing. Now let's break out some equations.
 Gravitational potential energy:
 E = m*g*h
 Kinetic energy:
 E = 0.5 m*v^2
   Since we know they need to be equal based upon the discussion above, let's set the equations equal to each other, solve for h, the substitute the known values and calculate.
 m*g*h = 0.5 m*v^2
 g*h = 0.5 v^2 h = 0.5 v^2/g
 h = 0.5*(3.6 m/s)^2 / (9.81 m/s^2)
 h = 0.5*12.96 m^2/s^2 / (9.81 m/s^2)
 h = 6.48 m^2/s^2 / (9.81 m/s^2)
 h = 0.660550459 m
   Rounding to 2 significant figures gives an initial height of 0.66 m.
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