Answer:
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[tex]_A[/tex] = E[tex]_B[/tex] = [tex]\frac{1}{2}[/tex]k( Δx )²
we substitute
= [tex]\frac{1}{2}[/tex] × k( 15 - 5 )² × 10⁻⁴
= [tex]\frac{1}{2}[/tex] × 605 × ( 10 )² × 10⁻⁴
= [tex]\frac{1}{2}[/tex] × 605 × 100 × 10⁻⁴
= 3.025 J
Therefore, the energy in the oscillations between the blocks is 3.025 J
