Question

The bob of a pendulum is released from a horizontalocal position.. if the length of pendulum is 1.5 m . What is speed of bob dissipated 5%of it's initial energy against air resistance

Answer 1

Length of the pendulum, l = 1.5 m

Mass of the bob = m

Energy dissipated = 5%

According to the law of conservation of energy, the total energy of the system remains constant.

At the horizontal position:

Potential energy of the bob, EP = mgl

Kinetic energy of the bob, EK = 0

Total energy = mgl … (i)

At the lowermost point (mean position):

Potential energy of the bob, EP = 0

Kinetic energy of the bob,EK=1/2mv²

Total energy… Ex=1/2mv²(ii)

As the bob moves from the horizontal position to the lowermost point, 5% of its energy gets dissipated.

The total energy at the lowermost point is equal to 95% of the total energy at the horizontal point, i.e.,

1/2mv²=95/100*mgl

Therefore,
v=√(2*95*1.5*9.8)/100
=5.28m/s

Answer 2

Answer:

Length of the pendulum, l = 1.5 m

Mass of the bob = m

Energy dissipated = 5%

According to the law of conservation of energy, the total energy of the system remains constant.

At the horizontal position:

Potential energy of the bob, EP = mgl

Kinetic energy of the bob, EK = 0

Total energy = mgl … (i)

At the lowermost point (mean position):

Potential energy of the bob, EP = 0

Kinetic energy of the bob, EK = (1/2)mv2

Total energy Ex = (1/2)mv2 ....(ii)

As the bob moves from the horizontal position to the lowermost point, 5% of its energy gets dissipated.

The total energy at the lowermost point is equal to 95% of the total energy at the horizontal point, i.e.,

(1/2)mv2 = (95/100) mgl

∴ v = (2 × 95 × 1.5 × 9.8 / 100)1/2

= 5.28 m/s