Spring Potential Energy Question?
The spring in a spring gun has a spring constant of 30.0 N/m. The spring gun is oriented so that it will fire horizontally. It is compressed 10.0 cm and fires a projectile with a mass of 50.0 g. What is the speed of the projectile as it leaves the spring gun?
Thanks for your help! Please include steps, stuck trying to work it out.
Answers & Comments
Verified answer
Use conservation of mechanical energy. Initial PE + KE equals final PE + KE
For PE in a spring, use PE = .5kx^2 where k is the spring constant and x is the spring's displacement. After the spring is activated, the PE turns into KE, which is shown as .5mv^2.
This can be expressed as (make sure to convert to proper units!)
.5kx^2 = .5mv^2
.5(30)(.1)^2 = .5(.05)(v)^2 <- Convert to meters, kg. Multiply by 2 to remove .5.
30(.01) = .05v^2
6 = v^2 <-- Multiply 30 and .01, divide both sides by .05
v = 2.45 m/s