A4.39\times10^5~\text{kg}4.39×10 5 kg subway train is brought to a stop from a speed of 1.08 miles per hour in 0.397 m by a large spring bumper at the end of its track. what is the force constant k of the spring?

6.5 x 10⁵ N/m

Explanation:

m = mass of the train = 4.39 x 10⁵ kg

v = speed of the train = 1.08 mph = 1.08 x 0.45 m/s = 0.483 m/s          (Since 1 mph = 0.45 m/s)

k = spring constant of the spring = ?

x = Compression of the spring = 0.397 m

Using conservation of energy

Spring energy gained due to compression of spring = Kinetic energy lost by train

(0.5) k x² = (0.5) m v²

k x² = m v²

k (0.397)² = (4.39 x 10⁵) (0.483)²

k = 6.5 x 10⁵ N/m

98.4 m

explanation:

let s = distance up the plane  

then by conservation of energy  

1/2m(18.3)^2 = mg s sin(10)  

(18.3)^2 = 2g s sin(10)  

s = (18.3)^2/2gsin(10)  

= 98.39564526  

= 98.4 m