A 70.0-g object connected to a spring with a force constant of 40.0 N/m oscillat

Question

A 70.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 6.00 cm on a frictionless, horizontal surface. (a) Find the total energy of the system. 72 Correct: Your answer is correct. mJ (b) Find the speed of the object when its position is 1.30 cm. (Let 0 cm be the position of equilibrium.) Incorrect: Your answer is incorrect. This is the speed of the object when it passes through the equilibrium point. m/s (c) Find the kinetic energy when its position is 2.50 cm. mJ (d) Find the potential energy when its position is 2.50 cm. mJ

Explanation / Answer

A) Total energy is E = 0.5*k*A^2 = 0.5*40*0.06^2 =0.072 J = 72 mJ

B) speed at 1.3 cm is v = w*sqrt(A^2-x^2) = sqrt(k/m)*sqrt(A^2-x^2) = sqrt(40/0.070)*sqrt(0.06^2-0.013^2) = 1.4 m/s

C) speed of the object when it passes through equilibrium point is V = A*w = 0.06*sqrt(k/m) = 0.06*sqrt(40/0.070) = 1.43 m/s

d) KE = 0.5*K*(A^2-x^2) = 0.5*sqrt(40/0.07)*(0.06^2-0.025^2) = 0.0355 J = 35.5 mJ

e) E = KE+PE

PE = E-KE = 72-35.5 = 36.5 mJ

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