The length of a pendulum is 25 cm, and the mass of the bob is 300 g. It is displ

Question

The length of a pendulum is 25 cm, and the mass of the bob is 300 g. It is displaced from the equilibrium position and released. At the lowest point of the trajectory the KE of the pendelum is 0.265 J.

A. How fast is the pendelum moving at the lowest point?

B. From what height (above the lowest point of the trajectory) was it released?

C. What is the total mechanical energy of the pendulum?

D. The pendulum is passing through a point 5 cm above its equlibrium position (its lowest point).

- find its PE (Potential Energy)

- find its KE (Kinetic Energy)

No answer choices sorry!!! Thanks for your time

Explanation / Answer

A )1/2mv^2= k.E.

1/2×0.3×v^2=0.265

v=1.33m/s

B)acc to law of conservation of energy

mgh=0.265

h=0.09m

C) total mechanical energy=0.265J

D) mgh=0.3×9.8×0.05=0.147J

Acc to law of conservation of energy= mgh+1/2mv^2=0.265

1/2mv^2=0.265-0.147=0.118J

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