Spring constant???? Physics help, please!!!

[From: ] [author: ] [Date: 11-12-05] [Hit: ]

81 * time^2Time = (1.9 ÷ 4.905)^0.5 Horizontal distance = horizontal velocity * time4.41 = horizontal velocity * (1.9 ÷ 4.......

The block flies thru the air & falls vertically 1.9 m in the time it takes to travel 4.41 m horizontally. From this & the equations I don't remember, you can find the blocks vertical speed as it leaves the track. Probably from there it's a momentum equation to find the force & that allows you to work back to the spring constant.

But I may be wrong.

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H=1.9 m
distance traveled=4.41 m

As the spring expands, the block accelerates to its maximum horizontal velocity. As the potential energy of the spring decreases, the kinetic energy of the block increases to maximum amount. The horizontal velocity remains constant as the block falls to the ground. As the block falls, its vertical velocity increases 9.81 m/s each second.

Now let’s go backwards and determine the time that the block was falling. Then determine the horizontal velocity of the block as it left the track. Then determine the kinetic energy of the block, then potential of the spring; and finally,k!

Vertical distance = ½ * 9.81 * time^2
1.9 = ½ * 9.81 * time^2
Time = (1.9 ÷ 4.905)^0.5

Horizontal distance = horizontal velocity * time
4.41 = horizontal velocity * (1.9 ÷ 4.905)^0.5
horizontal velocity = 4.41 ÷ (1.9 ÷ 4.905)^0.5

The kinetic energy of the block = ½ * 0.528 * [4.41 ÷ (1.9 ÷ 4.905)^0.5 ]^2
Potential energy of the spring = ½ * k * 0.047^2
½ * k * 0.047^2 = ½ * 0.528 * [4.41 ÷ (1.9 ÷ 4.905)^0.5 ]^2
Divide both sides by ½ * 0.047^2

k = 0.528 * [4.41 ÷ (1.9 ÷ 4.905)^0.5 ]^2 ÷ 0.047^2

Solve for k = 12,000.54201 N/m
Round as your instructor desires.

I hope this helps you understand the process!

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