a 2kg mass is suspeneded from the bottom of a spring A 2 kg mass is suspended fr

Question

a 2kg mass is suspeneded from the bottom of a spring

A 2 kg mass is suspended from the bottom of a spring (k = 400N/m). Initially the spring is at rest. You pull the mass down an additional 5cm to x = -5cm and release it at time t = 0. What is the force of the spring on the mass at time t = 0? Write an expression for X(t), V(t) and a(t) for t > 0. What are the values of the kinetic and potential energies of the system when x = 0.04m? What position, velocity and time will the mass have the first time that exactly half of the energy of the system is in the form of kinetic energy?

Explanation / Answer

a) F = kx = 400*.05= 20 N

b) x = A cos( w t)

A = -0.05

w = sqrt(k/m) = sqrt(400/2)=14.14

so x = -0.05cos(14.14 t)

v = A w sin(wt) = -0.707 sin(14.14 t)

a = -A w^2 cos(wt) = 10*cos(14.14 t)

c) PE = 1/2 kx^2 = 0.5*400*0.04^2= 0.32 m

KE = E - PE = 1/2 k(A^2 - x^2) =0.5*400*(0.05^2-0.04^2)= 0.18 J

d)

KE = 1/2 E so

PE = 1/2 E

1/2 k x^2 = 1/4 k A^2
x = A/sqrt(2) = -0.05/sqrt(2)=-0.0353

then 1/2 mv^2 = 1/2 E
2*v^2 = 0.5*400*0.05^2
v=0.5 m/s

use that x = A cos(wt)
-0.0353 = -0.05*cos(14.14*t)

t=0.0557 s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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