The force constant of a spring is 600 N/m andthe unstretched length is 0.72 m. A

Question

The force constant of a spring is 600 N/m andthe unstretched length is 0.72 m. A 3.2-kg block is suspended fromthe spring. An external force slowly pulls the block down, untilthe spring has been stretched to a length of 0.86 m. The externalforce is then removed, and the block rises. In this situation, whenthe spring has contracted to a length of 0.72 m, what is the upwardvelocity of the block? The force constant of a spring is 600 N/m andthe unstretched length is 0.72 m. A 3.2-kg block is suspended fromthe spring. An external force slowly pulls the block down, untilthe spring has been stretched to a length of 0.86 m. The externalforce is then removed, and the block rises. In this situation, whenthe spring has contracted to a length of 0.72 m, what is the upwardvelocity of the block?

Explanation / Answer

the spring is stretched by x= 0.86-0.72=0.14 m k=force constant elastic potential energy gain by the spring=(1/2)*k*x2=(1/2)*600*0.142=5.88 J this elastic potential energy is converted to the kinetic energy ofthe block m=3.2 let velocity =v (1/2)*m*v2=5.88 (1/2)*3.2*v2=5.88 v=1.92 m/s upward velocity of the block=1.92 m/s (ans)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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