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iogann1982 [59]
3 years ago
9

A 160 g air-track glider is attached to a spring. The glider is pushed in 11.2 cm and released. A student with a stopwatch finds

that 14.0 oscillations take 19.0 s . You may want to review (Pages 400 - 402) . For help with math skills, you may want to review: Solving Radical Equations For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Mass on a spring. Part A What is the spring constant
Physics
1 answer:
Yanka [14]3 years ago
6 0

Answer:

k = 3.41 N/m

Explanation:

The time period is given as:

T = \frac{time\ taken}{No.\ of\ oscillations} \\\\T = \frac{19\ s}{14} \\\\T = 1.36\ s

Another formula for the time period of the spring-mass system is:

T = 2\pi\sqrt{\frac{m}{k}} \\\\(1.36\ s)^2 = 4\pi^2\frac{0.16\ kg}{k}\\\\k = \frac{(4\pi^2)(0.16\ kg)}{(1.36\ s)^2}\\\\

<u>k = 3.41 N/m</u>

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