If a body under the action of a braking force of 2.1 N passes a path of 38 m to a complete stop

| September 11, 2021 | education

If a body under the action of a braking force of 2.1 N passes a path of 38 m to a complete stop, then its kinetic energy at the moment of the onset of braking is?

If a body under the action of a braking force of 2.1 N passes a path of 38 m to a complete stop, then its kinetic energy at the moment of the onset of braking is? F = 2.1 N. S = 38 m. V = 0 m / s. Ek -? The kinetic energy of the body, Ek, is determined by the formula: Ek = m * V ^ 2/2, where m is the mass of the body, V is the speed of the body. At the moment of the beginning of deceleration, the speed of the body was Vо, therefore Ek = m * Vо ^ 2/2. Let’s write Newton’s 2 law: F = m * a. a = (V ^ 2 -Vo ^ 2) / 2 * S, since V = 0 m / s, then a = -Vo ^ 2/2 * S. The sign “-” means that the speed and acceleration are directed in opposite directions … a = Vo ^ 2/2 * S. F = m * Vo ^ 2/2 * S. m * Vo ^ 2/2 = F / S. Ek = F / S. Ek = 2.1 N / 38 m = 0.055 J.
Answer: kinetic energy at the beginning of deceleration Ek = 0.055 J.



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