Question

A simple pendulum consists of a 4.9-kg mass attached to a string. The pendulum is pulled to the right and held at rest so that it is 2.08 m above the lowest point in its swing. If you let it go from this point, how fast (in m/s) is it traveling at the lowest point of its swing (on the initial pass)

Answer 1

6.384 m/
`sec^(2)`

Step-by-step explanation:

The velocity is given by;

We know;

PE= mgh

KE= 1/2 m
`V^(2)`

but KE=PE

==> 1/2 m
`V^(2)` =mgh

==> 1/2
`V^(2)` =gh

==>
`v=√(2gh)`

puting g=9.8 m/sec2

h=2.08 m

==>
`v= 6.384 m/sec^(2)`=