Question

A 0.45 kg object rests on a frictionless horizontal surface, where it is attached to a massless spring whose k-value equals 24.0 N/m. Let x be the displacement, where x = 0 is the equilibrium position and x > 0 when the spring is stretched. The object is pushed, and the spring compressed, until x, =-4.00 cm. It then is released from rest and undergoes simple harmonic motion. (a) What is the object's maximum speed (in m/s) after it is released? 29 m/s (b) How fast is the object moving (in m/s) when the spring is momentarily compressed by 1.80 cm (that is, when x-1.80 cm)? 16 Mechanical energy is conserved in this system, and the gravitational term remains unchanged (since all motion is horizontal). Write an expression for mechanical energy that includes the kinetic energy and the potential energy of the spring. Which term(s) can be ignored when the spring is compressed and the object at rest? Use the remaining terms, and the given quantities, to solve for the maximum speed. m/s (c) How fast is the object moving (in m/s) whenever the spring is extended by 1.80 cm (that is, when passing through x = +1.80 cm)? 16 You are computing the speed of the object when it has the same distance from the midpoint as it had in part (b)-but on the other side. Does any term in the energy-conservation equation depend on the sign of the displacement, x? Could this insight help you answer this question once you have correctly solved part (b)? (d) Find the magnitude of the displacement (in cm) at which the object moves with one-half of the maximum speed 1x1 = 4 Mechanical energy is conserved in this system, and the gravitational term remains unchanged (since all motion is horizontal). Write an expression for mechanical energy that includes the kinetic energy and the potential energy of the spring. Which term(s) can be ignored when the spring is compressed and the object at rest? Use the remaining terms, and the given quantities, to solve for the displacement that yields half of the maximum speed. (You should have an expression and value for the maximum speed from having solved part (a) above.) cm

Answer 1

Tofal enn at х, 1.ecm 2 2 2 2 2 released Kon ret 24. Г 16-324 0.45 m/s m 53-33 X 12, X 2) V, - 26.09 ms