You exert a 138 N push the leftmost of two identical blocks of mass 244 g connected by a spring of stiffness 605 kg/s2. After pushing the block a distance 15 cm, you release it; by this time the rightmost block has moved a distance 5 cm. (a) What is the energy in the oscillations between the blocks

the energy in the oscillations between the blocks is 3.025 J

Explanation:

Given the data in the question;

Force f = 138 N

stiffness of spring k = 605 kg/s²

mass of block = 202 g = 0.202 kg

pushing the block a distance 15 cm, the rightmost block has moved a distance 5 cm

i.e

x₁ = 15 cm

x₂ = 5cm

the energy in the oscillations between the blocks will be;

E = E = k( Δx )²

we substitute

= × k( 15 - 5 )² × 10⁻⁴

= × 605 × ( 10 )² × 10⁻⁴

= × 605 × 100 × 10⁻⁴

= 3.025 J

Therefore, the energy in the oscillations between the blocks is 3.025 J

answer; i believe the correct answer is (40,000 j) good luck

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pour in room temperature or cold water. cover the pitcher and refrigerate. steep white or green tea for 6 to 8 hours; steep black or oolong tea for 8 to 12 hours.jun 28, 2015