# Working with two springs placed in series.

a 3-lb block can slide without friction in a slot and is connected to two springs of constant k_1=80 lb/ft and k_2=6 lb/ft. The springs are initially unstreatched when the block is pulled 2 in. to the right and released.Determine (a) the maximum velocity of the block, (b)the velocity of the block when it is 0.8 in. from its initial position.

See attached for diagram.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

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Consider two springs placed in series with a mass on the bottom of the second. The force is the same on each of the two springs. Therefore

(1)

Solving for in terms of ,

(2)

We are looking for the effective spring constant so that

(3)

Where is the total displacement of the mass. Equating (3) with the right side of (1) and substituting into (2) gives

(4)

Divide through by to obtain

(5)

We therefore have

(6)

Thus the effective spring constant is K = k 1 k 2 /(k 1 + k 2) = 80*60/(80+60) = 4800/140 = 34.28 lb/ft

We have,

1 INCH = 2.54 CENTIMETERS

1 FOOT = 30.48 CENTIMETERS

2 inches = 2*2.54 cm = 5.08 cm = 5.08/30.48 ft = 0.167 ft

The energy stored in a stretched/compressed spring is E = (1/2) k x^2

Here E = 0.5 * 34.28 * (0.167)^2 = 0.47801746 lb ft

This much energy will get converted to the kinetic energy of the mass when the spring is released.

The max KE the mass can attain is (1/2) k x^2

Thus we can write, 0.47801746 = (1/2) * m v^2

With m = 3 lb, we get v^2 = 0.47801746 * 2/3 = 0.319

Or v = 0.564 ft/sec

We can write the total energy at point as the sum of the potential energy and the kinetic energy at that point since the total energy of the system is a constant.

Thus, Total energy = KE + PE = Â½ m v^2 + Â½ k x^2

Here total enegy = 0.478 lb ft = Â½ m v^2 + Â½ k x1^2

We have, 0.8 inches = 2.54*0.8/30.48 = 0.067 ft

= Â½ * 3 * v^2 + 0.5 * 34.28 * 0.067^2 = (3/2) V^2 + 0.077

 (3/2) v^2 = 0.478 - 0.077 = 0.401

 v^2 = 0.401*2/3 = 0.267

or v = 0.517 ft/s

we can see that the velocity at 0.8in is approaching the maximum value.

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