Physics Help please?

Question

A pendulum bob is released from a position 0.25m above the ground. What is the speed of the pendulum bob as it passes through the equilbrium position(past the ground)?

Answer 1

since the bob was intially at rest, at a height of 0.25m above the ground, then it had an Intial Potential Energy ( PEi) , right?

since it was intially at rest, the intial velocity =0m/s which means the intial Kinetic Energy = 0

when it reaches the equililbrium position, the height=0 , therefore the PE final = 0 and all the energy was transferred to the KE final.

so :

PE i = KE final

mgh=1/2 *m v(final) ^2

m cancel each other so

V final= sqrt (2gh)

Vf=v=2.21 m/s

Answer 2

Apply the law of conservation of energy to this problem.

Potential energy of bob = Kinetic energy of bob

In other words, as the bob is in the 0.25 m position (above the ground), it has potential energy. This is converted to kinetic energy (which is maximum) at the equilibrium position.

Potential energy = PE = mgh

Kinetic energy  = KE = (1/2)mV^2

where

m = mass of the pendulum bob

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

h = initial height of bob from the ground = 0.25 m (given)

V = velocity of the bob at just passed the equilibrium point



Substituting appropriate values,

m(9.8)(0.25) = (1/2)mV^2

and since "m" appears on both sides of the equation, it will simply cancel out.

Therefore,

(9.8)(0.25) = (1/2)V^2

V^2 = 4.9

V = 2.21 m/sec.