Respuesta :
Answer:
(a) the constant of the spring is 27095.375 J/m2
(b) 0.052 m/s
(c) F=17341.04 N
Explanation:
Hello
The law of conservation of energy states that the total amount of energy in any isolated system (without interaction with any other system) remains unchanged over time.
if we assume that there is no friction, then, the kinetic energy of the train (due to movement) will be equal to the energy accumulated in the spring
Step 1
energy of the train (kinetic)
[tex]E_{k}=\frac{m*v^{2} }{2}\\ E_{k}=\frac{2.15*10^{5}kg*(0.71\frac{m}{s}) ^{2} }{2}\\E_{k}=54190.75 J\\ \\[/tex]
step 2
energy of the spring
[tex]E_{s}=\frac{Kx^{2} }{2}\\[/tex]
where K is the constant of the spring and x the length compressed.
[tex]E_{s}=\frac{Kx^{2} }{2}=54190.75 J\\\frac{Kx^{2} }{2}=54190.75 J\\\\k=\frac{2*54190.75 J}{x^{2}}\\k=\frac{2*54190.75 J}{(2.0 m)^{2} }\\ k=27095.375\frac{J}{m^{2} } \\\\[/tex]
(a) the constant of the spring is 27095.375 J/m2
(b)
[tex]x=0.640 m\\\\E_{s}=\frac{27095.375 (\frac{j}{m^{2} })*(0.640m)^{2} }{2}\\ E_{s}=5549.1328 J\\[/tex]
equal to train energy
[tex]E_{k}=\frac{2.15*10^{5}kg*(v) ^{2} }{2}\\\frac{2.15*10^{5}kg*(v) ^{2} }{2}=5549.1328 J\\v^{2}=\frac{2*5549.1328 J}{2.15*10^{5}kg}\\v=0.052 \frac{m}{s} \\\\[/tex]
(b) 0.052 m/s
(c)
[tex]F= kx\\\\F=27095.375 \frac{J}{m^{2} }*(0.640 m)\\ F=17341.04 N\\F=17341.04[/tex]
(c) F=17341.04 N
I hope it helps
