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if KE=1/2mv2, then, a moving 3 times as fast as another will require how much work to come to a stop?

User Gcaprio
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The work required to stop a faster moving object is 9/2 times the work required to stop a slower moving object.

The equation for kinetic energy is KE = 1/2 * mv², where m is the mass of the object and v is its velocity.

If one object is moving 3 times as fast as another, then its velocity will be 3v, where v is the velocity of the slower object.

To calculate the work required to stop the faster object, we first need to find its kinetic energy.

Using the equation for kinetic energy, the kinetic energy of the faster object is KE = 1/2 * m(3v)² = 9/2 * mv².

Therefore, the work required to stop the faster object is 9/2 times the work required to stop the slower object.

User Aysebilgegunduz
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