Question Detail
Question
In Example 5, how fast is the block going when halfway to its maximum height?
Correct Answer
Option C is the correct answer.
Detailed Explanation
At halfway, h/2 = 0.64 m. Using energy conservation: ½kx_i² = ½mv_f² + mg(h/2). Solving for v_f gives v_f = √((k/m)x_i² - gh) = √((625/0.5)×0.01 - 9.8×0.64) = √(12.5 - 6.27) = √6.23 ≈ 2.5 m/s.
Hint
Energy is split between kinetic and potential at the halfway point.