Answer :
Complete question is;
A 5 × 10^(5) kg subway train is brought to a stop from a speed of 0.5 m/s in 0.4 m by a large spring bumper at the end of its track. what is the force constant k of the spring
Answer:
781250 N/m
Explanation:
From conservation of energy, potential energy is equal to kinetic energy.
Thus;
½mv² = ½kx²
where;
m = mass of train
v = velocity of train
k = force constant of spring
x = the distance the train went while being stopped
We are given;
Mass; m = 5 × 10^(5) kg
Velocity; v = 0.5 m/s
Distance; x = 0.4 m
Thus, from ½mv² = ½kx²
Divide both sides by ½ to get;
mv² = kx²
k = mv²/x²
k = [(5 × 10^(5)) × 0.5²]/0.4²
k = 781250 N/m