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]
