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A 2-kilogram block is dropped from a height of 0.45 meter an uncompressed spring, as shown. The spring has an elastic constant of 200 newtons per meter and negligible mass. The block strikes the end of the spring and sticks to it.

 

a. Determine the speed of the block at the instant it hits the end of the spring. 

b. Determine the force in the spring when the block reaches the equilibrium position 

c. Determine the distance that the spring is compressed at the equilibrium position 

d. Determine the speed of the block at the equilibrium position 

e. Is the speed of the block a maximum at the equilibrium position, explain.

1 Answer

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(a) Apply energy conservation from start to top of spring using h = 0 as top of spring. 

U = K

mgh = ½mv2

(9.8)(0.45) = ½v2

v = 3m/s

(b) At equilibrium the forces are balances 

Fnet = 0

Fsp = mg =(2)(9.8) = 19.6 N

(c) Using the force from part b, 

Fsp = k∆x

19.6 = 200∆x 

∆x = 0.098m

(d) Apply energy conservation using the equilibrium position as h = 0. 

(Note that the height at the start position is now increased by the amount of ∆x found in part c hnew = h + ∆x = 0.45 + 0.098 = 0.548m)

Utop = Usp + K 

mgh = ½k∆x2 + ½ mv2 

(2)(9.8)(0.548) = ½ (200)(0.098)2 + ½ (2)(v2

v = 3.13m/s

(e) This is the maximum speed because this was the point when the spring force and weight were equal to each other and the acceleration was zero. Past this point, the spring force will increase above the value of gravity causing an upwards acceleration which will slow the box down until it reaches its maximum compression and stops momentarily.

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