Physics Question

Question

A 1.2KG block is held against a spring of force constant 1.0x10⁴N/m, copressing it a distance of 0.15m.  How fast is the block moving after it is released and the spring pushes it away?

Answer 1

Use conservation of energy. 

Initial energy = final energy

potential spring = kinetic energy

1/2 k x² = 1/2 m v²

v² = k x² / m

v = sqrt[k x² / m] = sqrt[1.0e4N/m (0.15m)2 / (1.2kg)]

v = 13.7 m/s

Answer 2

This is a conservation of energy question. The block has a certain amount of potential energy before it is released, due to the compressed spring. This potential energy is equal to 1/2 k x^2. In this case, k (the spring constant) is 1.0x10^4 N/m and x (the spring's displacement) is 0.15m. This gives us 1/2*(1x10^4)*(0.15^2)=112.5 Joules. After it is released, all of this energy is transformed into kinetic energy, given by 1/2 m v^2. So, 112.5 J = 1/2 (1.2kg) v^2. Solving for v gives 13.7 m/s.