A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m.

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

(A) 3.7 m/s
(B) 0.18 m/s
(C) 5.2 m/s
(D) 13 m/s
(E) 0.13 m/s

Option C is the Correct answer. 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 5.2 m/s.

Explanation:

PE(max) = KE +PE

= `"kx"("max")^2/2` =`"mv"^2/2` + `"kx"^2/2`

= `"x(max)"` = `\sqrt(mv^2/k +x^2)`

= `\sqrt(0.25•9/200 +0.15^2)`= 0.184.

v(max) = ω•x(max) = x(max) •sqrt(k/m)

=0.184•sqrt(200/0.25)

=5.2 m/s