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 = 37.7 J and a maximum displacement from equilibrium of 0.220 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 = 0.220 m on a rough surface so that it loses 16.8 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 total mechanical energy            E = 37.7 J Amplitude is A = 0.220m a ) spring constant is 1/2 k A^2 = 37.7 J                                   1/2 k ( 0.220m)^2 = 37.7 J                                    k = 1.557*10^3 N /m b ) At the equilibrium point the potential energy of the system is zero so the total energy of the system is       K.E = 37.7 J c ) Given that maximum speed of the block is v = 3.45 m/s                 1/2 m v^2 = 37.7 J                  1/2 * m ( 3.45m/s)^2 = 37.7 J                      m = 6.33 kg d ) displacement is x = 0.160m        speed of the block is v = (k /m) (A^2 - x^2 )                                             = 1.557*10^3 N /m / 6.33 kg [ (0.220m)^2 - ( 0.160m)^2 ]                                            = 2.38 m/s e ) kinetic energy at x = 0.160m                K.E = 1/2 m v^2                        = 1/2 * 6.33 ( 2.38 m/s)^2                        = 17.9 J f ) Potential energy stored at x = 0.160m           P.E =1/2 k x^2                  = 1/2 *  1.557*10^3 N /m ( 0.160m)^2                  = 19.84J g ) The difference in the energy is equal to Potential energy stored in the spring               - 1/2 k x^2 = 37.7J -19.48 J                 - 1/2 *1.557*10^3 N /m * x = 17.86J                         x = - 0.151m                 

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