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Sagot :
Answer:
37.5 seconds
Explanation:
The given parameters are;
The mass of the block on the spring, m = 2.50 kg
The force with which the loaded spring launches the block, F = 450 N
The change in momentum of the block, Δp = 12.0 kg·m/s
We have;
Let the force with which the block was launched = The net force, [tex]F_{NET}[/tex]
By Newton's second law of motion, we have;
F = [tex]F_{NET}[/tex] = Δp × Δt
Where;
Δt = The time the block is in contact with the spring
Therefore;
[tex]\Delta t = \dfrac{F_{NET}}{\Delta p}[/tex]
By plugging in the values for [tex]F_{NET}[/tex] and Δp, we have;
[tex]\Delta t = \dfrac{450 \ N}{12.0 \ kg \cdot m/s} = 37.5 \ s[/tex]
The time duration the block is in contact with the spring, Δt = 37.5 seconds.
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