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 = 54.3 J and a maximum displacement from equilibrium 0.250 m. (a) What is the spring constant? 1737.6 N/m (b) What is the kinetic energy of the system at the equilibrium point? 54.3 J (c) If the maximum speed of the block is 3.45 m/s, what is its mass? 9.12 kg (d) What is the speed of the block when its displacement is 0.160 m? 2.65 m/s (e) Find the kinetic energy of the block at x = 0.160 m. 32.02 J (f) Find the potential energy stored in the spring when x = 0.160 m. 22.3 J (g) Suppose the same system is released from rest at x = 0.250 m on a rough surface so that it loses 14.8 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 ME = 54.3 J
X Max=0.25 m

so, K = TME * 2 / x2 =1737.6

total mechanical energy = constant

So,
KE + PE + Energy lost = TME

=> 0 + P.E. + 14.8 = 54.3 J
or,
P.E. = 39.5 J

The P.E. of spring = (1/2)(k)(x2)
or,
x = (39.5* 2/1737.6 )0.5 = 0.213 m

Please do not forget to upvote the answer as it encourages us a lot.
Feel free to ask any question or inform about any miscalculations.
Happy Chegging.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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