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 = 42.7 J and a maximum displacement from equilibrium of 0.236 m. (a) What is the spring constant? N/m (b) What is the kinetic energy of the system at the equilibrium point? J (c) If the maximum speed of the block is 3.45 m/s, what is its mass? kg (d) What is the speed of the block when its displacement is 0.160 m? m/s (e) Find the kinetic energy of the block at x = 0.160 m. J (f) Find the potential energy stored in the spring when x = 0.160 m. J (g) Suppose the same system is released from rest at x = 0.236 m on a rough surface so that it loses 11.7 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant? m

Explanation / Answer

.5K*amplitude^2=total energy K=1533.32N/m b.it is the same as total energy=42.7 J c..5mv^2=total energy m=7.17kg d.(.5)(7.17v^2+1533*.16^2)=42.7 v=2.53m/s e.KE=.5*7.17^2.53^2=22.94J f.PE=Total energy-KE=19.76J g.by conservation of energy 11.7+.5*1533*x^2=42.7 x=.2m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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