A 1.00 kg object is attached to a spring and placed on a horizontal, smooth surf

Question

A 1.00 kg object is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of 21.0 N is required to hold the object at rest when it is pulled 0.200 m from its equilibrium position (the origin of the x axis). The object is now released from rest with an initial position of xi = 0.200 m, and it subsequently undergoes simple harmonic oscillations. Where does this maximum speed occur? x = (d) Find the maximum acceleration of the object. Where does it occur? x = (f) Find the speed of the object when its position is equal to one third of the maximum value (g) Find the acceleration of the object when its position is equal to one third of the maximum value.

Explanation / Answer

(a) By F = -kx [-ve is just indicating the direction of Force] =>21= k x (0.2)^2 =>k = 525 N/m (b) By f = 1/2p x v[k/m] =>f = 1/(2 x 3.14) x v[525/1] =>f = 3.64 Hz (c) By PE(spring) = KE(spring) at the equilibrium position i.e. at the +0.20 m from the origion =>1/2mv^2 = 1/2kx^2 =>v^2 = 525 x (0.20)^2 =>v = v21 =>v(max) = 4.80 m/s (d) a(max) = -A?^2 [-ve just indicating the direction]at origin & +0.40 m position =>a(max) = 0.2 x (2 x p x f)^2 =>a(max) = 0.2 x (2 x 3.14 x 3.82)^2 =>a(max) = 115.10 m/s^2 (e) By PE(spring)[max] = 1/2kx^2 =>U = 1/2 x 525 x (0.20)^2 =>U = 10.50 J (f) By U = 1/2mv^2 + 1/2kx^2 =>10.5 = 1/2 x 1 x v^2 + 1/2 x 525 x (0.2/3)^2 (g) By a = -A?^2 =>a = -(0.2/3) x (2 x p x f)^2 =>a = -38.37 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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