A 0.38 kg object connected to a light spring with a force constantof 19.4 N/m os

Question

A 0.38 kg object connected to a light spring with a force constantof 19.4 N/m oscillates on a frictionless horizontal surface. If thespring is compressed 4.0 cm and released from rest. (a) Determine the maximum speed of theobject.
(b) Determine the speed of the object when the spring is compressed1.5 cm.
(c) Determine the speed of the object when the spring is stretched1.5 cm.
(d) For what value of x does the speed equal one-half themaximum speed?
(a) Determine the maximum speed of theobject.
(b) Determine the speed of the object when the spring is compressed1.5 cm.
(c) Determine the speed of the object when the spring is stretched1.5 cm.
(d) For what value of x does the speed equal one-half themaximum speed?

Explanation / Answer

                     Ep=kx2/2 = (19.4N/m*(0.04m)2/2 = 0.01552J when the speed at its max, Ep = 0; Kinetic energy = mv2/2 = 0.01552 J => v = 0.2858 m/s = 28.58 cm/s b) since energy conservation:        mv2/2 +k(x)2/2 = 0.01552 J =>     v = 0.2649 m/s = 26.49cm/s c) the same as (b):           v =0.2649 m/s = 26.49 cm/s d) if v = vmax/2 = 28.58 cm/s / 2= 14.29cm/s we have:           mv2/2 + k(x)2/2 = 0.01552 J =>x = 3.46 cm d) if v = vmax/2 = 28.58 cm/s / 2= 14.29cm/s we have:           mv2/2 + k(x)2/2 = 0.01552 J =>x = 3.46 cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---