A horizontal block-spring system with the block on a frictionless surface has to

Question

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 42.3 J and a maximum displacement from equilibrium of 0.259 m.

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 42.3 J and a maximum displacement from equilibrium of 0.259 m. What is the spring constant? What is the kinetic energy of the system at the equilibrium point? If the maximum speed of the block is 3.45 m/s, what is its mass? What is the speed of the block when its displacement is 0.160 m? Find the kinetic energy of the block at x = 0.160 m. Find the potential energy stored in the spring when x = 0.160 m. Suppose the same system is released from rest at x = 0.259 m on a rough surface so that it loses 16.2 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant?

Explanation / Answer

total energy left = 42.3-16.2 = 26.1 J

this is stored in spring. therefore, 1/2*k*x*x=26.1

26.1*2/1261.2=x*x

x= 0.203m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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