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 = 41.0 J and a maximum displacement from equilibrium of 0.201 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 = 0.201 m on a rough surface so that it loses 13.2 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

A horizontal block-spring system with the block on a frictionless surface has a total mechanical energy E=39J? and the maximum displacement from equilibrium of .200m. A. what is the spring constant? B. what is the KE of the system at the equilibrium point? C. if the max. speed of the block is 3.45 m/s what is its mass? D. what is the speed of the block when its displacement is 0.160m? E. find the KE of the block at x=.160m? F. find the PE stored in the spring when x=0.160m? G. suppose the same system is released from rest at x=0.240m on a rough surface so that it looses 14.0J by the time it reaches its first turning point(after passing equilibrium at x=0) what is its position at that instance? E=39 J Amplitude A = 0.200 m A)Let spring constant = k E = 1/2 *k*A^2 k = 2E/A^2 = 2*39/0.2^2 = 1950 N/m Ans: 1950 N/m B) At equilibrium, point kinetic energy = total mechanical energy = 39 J Ans: 39 J C) Let mass = M E = (1/2)*M*vmax^2 M = 2*E/vmax^2 = 2 * 39/3.45^2 = 6.55 kg Ans: 6.55 kg D)E = (1/2)Mv^2 + (1/2)kx^2 39 = (1/2)*6.55*v^2 + (1/2)*1950*0.16^2 39 = 3.28 v^2 + 24.96 3.28 v^2 = 14.04 v = sqrt(14.04/3.28) = 2.07 m/s Ans: 2.07 m/s E) KE = (1/2)Mv^2 = (1/2)*6.55*2.07^2 = 14 J Ans: 14 J F) PE = E-KE = 39 - 14 = 25 J Ans: 25 J G) PE1 = (1/2)*k*x1^2 = (1/2) * 1950 * 0.24^2 = 56.16 J KE1 = 0 KE2 = 0 x2 = ? Loss of energy = 14 J Therefore KE2 + PE2 = KE1 + PE1 - 14 0 + PE2 = 0 + 56.16 - 14 PE2 = 42.16 J (1/2)*k*x2^2 = 42.16 (1/2) * 1950 * x2^2 = 42.16 975 x2^2 = 42.16 x2 = sqrt(42.16/975) = 0.21 m Ans: 0.21 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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