A 0.454-kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 21.0 N/m. The
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Answer:
t = 1.41 sec.
Explanation:
If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.
First, we need to find the value of acceleration, which is the same for both blocks.
If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:
F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)
⇒ a = ( = g/5 m/s²
Once we got the value of a, we can use for instance this kinematic equation, and solve for t:
Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.