A spring with force constant k = 600, N/m is mounted as shown below. An 0.800-kg

Question

A spring with force constant k = 600, N/m is mounted as shown below. An 0.800-kg block attached to the end is undergoing simple harmonic motion with an amplitude of 0.0750 m There is no friction force on the block. x = 0 is defined to be the point of equilibrium. Compute (a) the maximum speed of the block; (b) the speed of the block when it is at x = 0.0300m; (c) the magnitude of the maximum acceleration of the block; (d) the acceleration of the block when it is at x = 0.0300m; (e) the total mechanical energy of the block at any point in its motion

Explanation / Answer

a) maximum speed is Vmax = A*w

A = 0.075 m

w = sqrt(k/m) = sqrt(600/0.8) = 27.4 rad/s

Vmax = A*w = 0.075*27.4 = 2.055 m/s

b) using law of conservation of energy

energy at x = 0 m = energy at x = 0.03m

0.5*m*vmax^2 = (0.5*m*v^2)+(0.5*k*x^2)

(0.5*0.8*2.055^2) = (0.5*0.8*v^2)+(0.5*600*0.03^2)

v = 1.88 m/sec

c) amax = w^2*A = 27.4^2*0.075 = 56.31 m/s^2

d) amax = w^2*x = 27.4^2*0.03 = 22.52 W/m^2

e) total mechanical energy is E = 0.5*k*A^2 = 0.5*600*0.075^2 = 1.69 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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