7. A spring of length of 25 cm is attached to a ceiling. A mass m = 0.80 kg is a

Question

7. A spring of length of 25 cm is attached to a ceiling. A mass m = 0.80 kg is attached to the spring and stretches to a total equilibrium length of L0 = 34 cm.
(a) Find the spring constant of the spring. (6 pts).
(b) You lift the mass until the spring returns to its original length (25 cm), and release it. When the mass returns to the equilibrium length, what is its speed? (6 pts).
(c) After you release the mass and it falls, what is the total length of the spring when the mass reaches its lowest point? (6 pts).
PLEASE INCLUDE ALL WORKOUT

Explanation / Answer

(a.)spring force balances weight of the mass.
let the spring constant be k.
then k*(34-25)*0.01=0.8*9.8
solving we get , k=87.11 N/m.
(b).the mass gains potential energy when it rises up and this energy gets converted to spring potential energy and kinetic energy when it falls.
so potential energy gained=0.09*0.8*9.8=0.7056 joule.
spring potential energy=0.5*k*x^2=0.5*87.11*0.09^2=0.3528 joule.
kinetic energy=0.7056-0.3528=0.3528 joule
speed=sqrt(2*energy/mass)=0.939 m/s.
(c).when it reaches the lowest point,it uses up the kinetic energy to convert to potential energy.
m*g*h=kinetic energy
solving for the distance h, h=0.045 m.

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

---

---

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