Asked by Huhu.

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
A3.7 m/s
B0.18 m/s
C5.2 m/s
D13 m/s
E0.13 m/s
Answered by Elena
PE(max) = KE +PE
kx(max)²/2 =mv²/2 +kx²/2
x(max) = sqrt(mv²/k +x²)=
=sqrt(0.25•9/200 +0.15²)=0.184.
v(max) = ω•x(max) = x(max) •sqrt(k/m) =
=0.184•sqrt(200/0.25)=5.2 m/s
Answered by Anonymous
A 0.25 kg block oscillates on the end of a spring with a spring constant of 100 N/m. If the
oscillation is started by elongating the spring 0.1 m and giving the block a speed of 3 m/s,
then the amplitude of the oscillation is:
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