Spring physics question
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Spring physics question

[From: ] [author: ] [Date: 11-12-03] [Hit: ]
6 m below the point at which it left the spring.What is the spring constant?Express your answer using two significant figures.thanks!!-Potential Energy of Spring = (1/2) k x^2 = (1/2) k (0.......
In a physics lab experiment, a spring clamped to the table shoots a 19 g ball horizontally. When the spring is compressed 22 cm, the ball travels horizontally 4.9 m and lands on the floor 1.6 m below the point at which it left the spring.

What is the spring constant?
Express your answer using two significant figures.


thanks!!

-
Potential Energy of Spring = (1/2) k x^2 = (1/2) k (0.22)^2
= Initial Kinetic Energy of Ball = (1/2) m v^2 = (1/2) (0.019) (Vh)^2
therefore, k = 0.3926 (Vh)^2 where Vh is the speed of the ball in the horizontal dimension.

To find Vh we must first determine the time that the ball is in the air for...
distance = (1/2) a t^2 where a = g = 9.8 m/s^2 and distance = 1.6 m
therefore, t = sqrt{ 2 * 1.6 / 9.8 } = 0.5714 seconds

Vh = distance / time = 4.9 / 0.5714 = 8.575 m/s

now we can use our previous equation!

k = 0.3926 (Vh)^2 = 0.3926 * (8.575) ^ 2

= 29
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