A horizontal spring-mass system oscillates on a frictionless table.If the ratio of the mass to the spring constant is 0.038 kg · m/N,and the maximum speed of the mass was measured to be 17.60 m/s,find the maximum extension of the spring.
A)3.4 m
B)3.4 cm
C)0.67 m
D)67 cm

Question

Quizplus · Accepted Answer

The maximum extension of the spring is given by:$$x_{max} = \frac{mv_{max}^2}{k}$$where:* m is the mass of the object* v_{max} is the maximum speed of the object* k is the spring constantSubstituting the given values into the equation, we get:$$x_{max} = \frac{(0.038 	ext{ kg})(17.60 	ext{ m/s})^2}{k}$$$$x_{max} = 3.4 	ext{ m}$$Therefore, the correct answer is A.

A Horizontal Spring-Mass System Oscillates on a Frictionless Table