A subway train is brought to a stop from a speed of 0.500 m/s in 0.400 m by a large spring bumper at the end of its track. What is the force constant of the s

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

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

f=ma newton

f=100x2.5=250n