Asimple pendulum, 2.0 m in length, is released with a push when the support string is at an angle of 25° from the vertical. if the initial speed of the suspended mass is 1.2 m/s when at the release point, what is its speed at the bottom of the swing? (g = 9.8 m/s2)

2.3m/s

Explanation:

mass-energy balance: ke(f) + pe(f) = ke(o) + pe(o)

since we are looking for the point at the bottom of the pendulum, thats the reference point, the lowest in the system. pe(f) is 0, since h

ke(f)=0.5m x v(f)^2

pe(f)=0

ke(o)=0.5m x v(o)^2

pe(o)-mxgxh

find h by: drawing a triangle with the pendulum at the vertical, then displaced by 25 degrees , The difference in height is h, because cos(25)=(adj)/(hyp)=(2-h)/2. I found h=0.187m

In the m-e balance, cancel the masses in all the terms.

.5xv(f)^2 =0.5v(o)^2 +gxh

Given v(o) = 1.2 m/s and g = 9.8 then v(f) = 2.2595 m/s

Therefore V(0) = 2.3 m/s

because if the tool that you are using is not balanced it there would be false mass.