A 0.25-kg object suspended on a light spring is released from aposition 5 cm abo

Question

A 0.25-kg object suspended on a light spring is released from aposition 5 cm above the stretched equilibrium position. The springhas a spring constant of 80 N/m.

(a) What is the total energy of the system? (Neglectgravitational potential energy).

(b) Does this energy depend on the mass of the object?Explain.

What is the speed of the object when the object is
(c) 5.0 cm above its equilibrium position and (d) 5.0 cm below itsequilibrium position?

(e) What is the object’s maximum speed, and where doesthis occur?

Explanation / Answer

Here the force applied on the mass is F=mg=0.25*g=2.45N We  the maximum displacement produced is F=kx x=2.45/80=0.030625m The total energy of the system is PE=1/2*k*x2 PE=0.0375J a. E=0.0375J b. Yes because its due to mass only the spring isstretched. c. E=PE+KE At equilibrium position KE=PE at extreme, 1/2*80*0.052=1/2*0.25*v2 v=0.8944m/s d. v=0.8944m/s e. v=0.894m/s This occurs at the equlibrium position of it. Hence we get by it. "Hope this helps!Best of luck for the rest of yourcoursework." v=0.8944m/s d. v=0.8944m/s e. v=0.894m/s This occurs at the equlibrium position of it. Hence we get by it. "Hope this helps!Best of luck for the rest of yourcoursework."

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