Answer:
See Below
Explanation:
Question:
A 200 g dart is pressed against the spring of a toy dart gun. The spring (with spring constant k = 500 N/s) is compressed by 6.0 cm and released. If the dart detaches from the spring when the spring reaches its equilibrium length (x = 0), what is the launch speed of the dart?
Answer:
PE = KE
1/2kx^2 = 1/2mv^2
Sole for v:
v = sqrt(kx^2/m)
v = sqrt(500*(0.06)^2/0.200)
v = sqrt(1.8/0.200
v = sqrt(9)
v = 3 m/s
I really hoped this helped you. :) Sorry if it didn't.