A block of mass m = 1.50 kg attached to a horizontal spring with force constant

Question

A block of mass m = 1.50 kg attached to a horizontal spring with force constant k = 6.00 times 10^2 N/m that is secured to a wall is stretched a distance of 5.00 cm beyond the spring's relaxed position and released from rest. What is the elastic potential energy of the block-spring system a. just before the block is released and b. when the block passes through the spring's relaxed position? What is the speed of the block c. as it passes through the spring's relaxed position and d. when it has compressed the spring 2.50 cm beyond its relaxed position?

Explanation / Answer

m= 1.50 kg

K = 600 N/m

dx = 0.05 m

a) Elastic potential Energy = 0.5*K*X^2

= 0.75 Joules.

B)

At relaxed position,

dx= 0; hense

Elastic potential Energy =0

C)

0.5x1.50xV^2 =0.75

V = 1m/s

D)

0.5*600*(0.025)^2 = 0.5*1.50*V'^2

V' = 0.5 m/s

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