K=100 н/м
l1=5см
F1=F2
k2-? l2=1 см

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вопрос

To determine the spring constant ( k_2 ) for the second spring, we can use Hooke's Law:

[ F = kx ]

Since the force acting on both the first and second springs are equal (( F_1 = F_2 )), we can set up the following equation:

[ k_1x_1 = k_2x_2 ]

Given that ( k_1 = 100 \, \text{N/m} ), ( x_1 = 5 \, \text{cm} = 0.05 \, \text{m} ), and ( x_2 = 1 \, \text{cm} = 0.01 \, \text{m} ), we can substitute these values into the equation and solve for ( k_2 ):

[ 100 \times 0.05 = k_2 \times 0.01 ]

[ 5 = 0.01k_2 ]

[ k_2 = \frac{5}{0.01} ]

[ k_2 = 500 \, \text{N/m} ]

Therefore, the spring constant for the second spring (( k_2 )) is 500 N/m.