A particle of mass m = 3 kg is released from
point A on a frictionless track at height hA =
6.28 m.
The acceleration of gravity is 9.8 m/s
2
.
Determine the particle’s speed at point B
at height hB = 2.2 m.
Answer in units of m/s

Question

anonymous · Answer

in this case, 
GPE = gravitational potential energy
the amount of kinetic energy = GPE_6.28m - GPE_2.2m
                                          = 3(6.28)(9.8) - 3(2.2)(9.8)
                                          = 120.0
KE = 0.5mv^2 = 120.0
therefore, speed = 120.0/(0.5)(3)
                         = 40m/s