A pendulum is pulled to a height of 1.28 m above the lowest position and released. What is its speed at the level of the lowest position?

Question

 possible answers include 

 

5.01 m/s
9.81 m/s
19.6 m/s
25.0 m/s
none of these

Answer 1

This is a typical energy conservation problem.  Set potential energy at height equal to kinetic energy at bottom.

 

mgh = (1/2)mv²

 

gh = v²/2

 

v = √(2gh) = √[2(9.8)(1.28)] m/s ≅ 5.01 m/s (rounded to 2 decimal places) 

 

Answer 2

When you lift the pendulum weight by 1.28 m, you increase its potential energy (PE) by:

 

      PE = mgh

• m = the pendulum weight's mass
• g = the acceleration due to gravity = 9.8 m/s2
• h = the height lifted = 1.28 m

 

When you release the pendulum, it returns to its lowest point converting the potential energy to kinetic energy, mv²/2:

 

      mgh = mv²/2

      Solve for v