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As a 15,000 kg jet lands on an aircraft carrier, its tail hook snags a cable to slow it down. The cable is attached to a spring with spring constant 60,000 N/m. If the spring stretches 30 m to stop the plane, what was the plane's landing speed?

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contexto1019

The plane's landing speed was 250 m/s.

The force required to slow the plane down is equal to the spring constant multiplied by the amount it stretched. Thus, we can use the equation:

  • F = kx

where F is the force required, k is the spring constant, and x is the amount stretched.

In this case:

  • F = 60,000 N/m * 30 m = 1,800,000 N

We can use the equation for momentum to solve for the plane's speed:

  • p = mv

where p is the momentum, m is the mass, and v is the velocity.

In this case:

  • p = 15,000 kg * v

Rearranging for v, we get:

  • v = p/m = 1,800,000 N/15,000 kg = 250 m/s

Learn more about Spring Constant here:

https://brainly.com/question/23885190

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