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 = 46.2 J and a maximum displacement from equilibrium of 0.260 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 = 0.260 m on a rough surface so that it loses 14.2 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

1/2*K*X^2=46.2 s0 1366.86=k b)E = 46.2 J c)1/2*m*v^2=46.2 m=7.76kg d)at x=1.6 p.e=.5*1366.86*1.6*1.6=1.749kj k.e=j=1/2mv^2 s0 27.19m/sec=v e)k.e=46.2kj

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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