Posted by Huhu. on .
A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the oscillation is started by elongating the spring 0.15 m and giving the block a speed of 3.0 m/s, then the maximum speed of the block is
PE(max) = KE +PE
kx(max)²/2 =mv²/2 +kx²/2
x(max) = sqrt(mv²/k +x²)=
v(max) = ω•x(max) = x(max) •sqrt(k/m) =