contestada

Four railroad cars, each of mass 2.50 3 104 kg, are coupled together and coasting along horizontal tracks at speed vi toward the south. A very strong but foolish movie actor, riding on the second car, uncouples the front car and gives it a big push, increasing its speed to 4.00 m/s southward. The remaining three cars continue moving south, now at 2.00 m/s. (a) Find the initial speed of the four cars. (b) By how much did the potential energy within the body of the actor change

Respuesta :

okpalawalter8

Answer:

a

  [tex]u = 2.5 \ m/s[/tex]

b

  [tex]\Delta PE = 28125 \ J[/tex]    

Explanation:

From the question we are told that

   The mass of the each railroad car is [tex]m = 2.50 *10^{4} \ kg[/tex]

    The magnitude of the increased speed of the first car is  [tex]v = 4.0 \ m/s[/tex]

    The magnitude of the speed of the other cars is  [tex]v_o = 2.0 \ m/s[/tex]

Generally from the law of linear momentum conservation is mathematically represented as

          [tex]4m * u = m * v +3m * v_o[/tex]   [initial momentum before the push = momentum after the  push ]

So

     [tex]4(2.50 * 10^4) * u = (2.50 * 10^4) *4 +3(2.50 * 10^4) *2[/tex]

=> [tex]u = 2.5 \ m/s[/tex]

Generally the change in potential energy in the actor is equivalent to the workdone by the actor which is equivalent to the change in kinetic energy of the  other three cart, this can be mathematically represented as

          [tex]\Delta PE =W = \Delta KE= \frac{1}{2} * m * u^2 - \frac{1}{2} * m * v_o^2[/tex]

=>      [tex]\Delta PE = \frac{1}{2} * 2.50*10^{4} * 2.5^2 - \frac{1}{2} * 2.5*10^{4} * 2^2[/tex]

=>      [tex]\Delta PE = 28125 \ J[/tex]        

   

Otras preguntas