Question

The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 2.15 x 10^5 kg train moving at 0.710 m/s. (a) What is the force constant of the spring? (N/m)
(b) What speed would the train be going if it only compressed the spring 0.640 m?
(c) What force does the spring exert when compressed 0.640 m? (Enter the magnitude only.)

Answer 1

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)

$E_{k}=\frac{m*v^{2} }{2}\\ E_{k}=\frac{2.15*10^{5}kg*(0.71\frac{m}{s}) ^{2} }{2}\\E_{k}=54190.75 J$

step 2

energy of the spring

$E_{s}=\frac{Kx^{2} }{2}$

where K is the constant of the spring and x the length compressed.

$E_{s}=\frac{Kx^{2} }{2}=54190.75 J\\\frac{Kx^{2} }{2}=54190.75 J\\\\k=\frac{2*54190.75 J}{x^{2}}\\k=\frac{2*54190.75 J}{(2.0 m)^{2} }\\ k=27095.375\frac{J}{m^{2} }$

(a) the constant of the spring is 27095.375 J/m2

(b)

$x=0.640 m\\\\E_{s}=\frac{27095.375 (\frac{j}{m^{2} })*(0.640m)^{2} }{2}\\ E_{s}=5549.1328 J$

equal to train energy

$E_{k}=\frac{2.15*10^{5}kg*(v) ^{2} }{2}\\\frac{2.15*10^{5}kg*(v) ^{2} }{2}=5549.1328 J\\v^{2}=\frac{2*5549.1328 J}{2.15*10^{5}kg}\\v=0.052 \frac{m}{s}$

(b) 0.052 m/s

(c)

$F= kx\\\\F=27095.375 \frac{J}{m^{2} }*(0.640 m)\\ F=17341.04 N\\F=17341.04$

(c) F=17341.04 N

I hope it helps