Answered

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A loaded spring launches a 2.50 kg block, using a force of 450 N. If the change in
momentum is 12.0 kg*m/s, how long was the block in contact with the spring?

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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