A 200 g block attached to a spring with spring constant 2.6 N/m oscillates horiz

Question

A 200 g block attached to a spring with spring constant 2.6 N/m oscillates horizontally on a frictionless table. Its velocity is 18 cm/s when x_0 = -5.8 cm. Significant Figures Feedback: Your answer 1.001 m/8^2 (=100.1 cm/s^2) was either differently or used a different number of significant figures than required for this part.What is the block's position when the acceleration is maximum Express your answer to two significant figures and include the appropriate units. What is the speed of the block when x_1 = 2.7 cm

Explanation / Answer

1) By PE(max) = KE(max) = 1/2kA^2
=>1/2mv^2 = 1/2kA^2
=>200 x 10^-3 x (0.18)^2 = 2.6 x A^2
=>A = [2.49 x 10^-3]
=>A = 4.99 x 10^-2 m OR 4.99 cm
2)By a = -A^2 {-ve just indicating the direction}
=>a = -A x [[k/m]^2
=>a = -Ak/m = [4.99 x 2.6 x 10^-2]/(200 x 10^-3) = 64.87 m/s^2
3)At the maximum stretched/compressed position.
4) The block will never be at position of 2.7 cm as it will oscilate between -5.8+/-4.99 = -10.79 & -0.81 mark.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.