Question

A Gotye Soi x V M Mail New M x Boy & Bear x S ANGEL Learni x Interactive Ca x V S Mail Inbox Cx LON-CAPA CC x D about blank x ON-CAPA Er x LON-CAPAN x LON-CAPA Cox R Arccos(x) calc x 3962f 7698 48bo 2823180S Q https: c/48/5a5923219f8c7041e 4e8c77089729208187a04 4324322 de756c075d5a4245a591d8 9f3e94189ed15f5c7b729e 14.dfdee72676ef3c6 Messages Course Help Log Joseph Raymond Zajac v Stud section: 730 PHY 233B Fall 2014 Main Menu Contents Grades Timer Notes Evaluate back Print L Course Contents HW 10 (11 20 Th 59 PM Energy, speed and acceleration of an oscillating mass A mass-Sprina system oscillates with an amplitude of 4.60 cm. If the spring constant is 223 N/m and the mass is 576 g, determine the mechanical energy of the system 1.14 J Submit Answer Incorrect. Tries 2/12 Previous Tries Determine the maximum speed of the object. Submit Answer Tries 0/12 ine the maximum acceleration. Submit Answer Tries 0/12 Send Feedback Post Discussion 7:51 PM 11/19/2014.

Answer 1

Ennery of the system is

E = (1/2) KA^2 = (1/2)(223) (0.0460 M)^2 = 0.235 j

maximum speed of the object is

v_max= omega A = square root k/m( A)= square root 223/0.576 kg (0.0460 m) = 0.90 m/s

maximum accleration is

a_max= omega^2 A = (k/m) A = 223/0.576 kg (0.0460 m)= 17.8 m/s^2

Answer 2

here w = sqrt(k/m) = sqrt(223/0.576) = 19.67

A) E = 0.5*m*w^2*A^2 = 0.5*0.576*19.67^2*0.046^2 = 0.235 J..

B) Vmax = A*w = 0.046*19.67 = 0.90482 m/sec..

C) amax = A*w^2 = 0.046*19.67^2 = 17.79 m/s^2