A 100kg crate is propelled uphill by a spring initially loaded with 400J of potential energy. the crate starts from rest. Point b is 0.2 m above point A. What is speed of the crate at point B?

Answer 1

initial energy = 400J
PE gained = mgh
total energy is conserved, so
KE = 400 - PE
1/2 mv^2 = KE, so solve for v

Answer 2

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