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A 1000 kg automobile is traveling at an initial speed of 20 m/s. It is brought to a complete stop in 5 s over a distance of 50 m. What is the work done in stopping the automobile

Respuesta :

cjrodriguez89

Answer:

W = -2*10⁵ J

Explanation:

  • Assuming no friction present, we can find the work done by an external force stopping the car, applying the work-energy theorem.
  • This theorem says that the total work done on one object by an external net force, is equal to the change in the kinetic energy of the object.
  • If the automobile is brought to a complete stop, we can find the change of the kinetic energy as follows:

        [tex]\Delta K = K_{f} - K_{0} = 0 - \frac{1}{2} * m* v_{0} ^{2} \\\\ \Delta K = -\frac{1}{2}*1000kg*(20 m/s)^{2} = -200000 J= -2e5 J[/tex]

  • So, the total work done in stopping the automobile, is -2*10⁵ J. The minus sign stems from the fact that the force and the displacement have opposite directions.
Cricetus

The work done will be "[tex]-2\times 10^5 \ J[/tex]".

Given values:

  • Mass, m = 1000 kg
  • Initial speed, v = 20 m/s
  • Distance, d = 50 m
  • Time, t = 5 s

As we know,

→ [tex]\Delta K = K_f - K_0[/tex]

           [tex]= 0-\frac{1}{2}mv_2^2[/tex]

By substituting the values,

           [tex]= - \frac{1}{2}\times 1000\times (20)^2[/tex]

           [tex]= - 500\times 400[/tex]

           [tex]= -200000 \ J[/tex]

           [tex]= -2\times 10^5 \ J[/tex]

Thus the above answer is right.

Learn more about speed here:

https://brainly.com/question/16794794

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