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 0.271m.

(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 = 0.271 m on a rough surface so that it loses 16.1 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant?
m

Can someone explain, D and down more in detail, becase I do not understand part D

Explanation / Answer

E = 42.3 J = 1/2 kA2

k = 1152 N/m

b) Kinetic energy of system at mean position = total energy =  E = 42.3 J

c) maximum speed of the block is 3.45 m/s, = A w

so w = 12.73

and k = mw2

so mass = 7.11 kg

d) v = w (A2 - x2 ) 1/2 = 2.77m/s

e) KE = 1/2 mv2 = 27.27 J( V= 2.77m/s)

F) PE =  E = 42.3 J - 27.27 J =15.03 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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