The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 1.10 ✕ 105 kg train moving at 0.350 m/s. (a) What is the force constant (in N/m) of the spring? N/m (b) What speed (in m/s) would the train be going if it only compressed the spring 0.600 m? m/s (c) What force (in N) does the spring exert when compressed 0.600 m? 2020 N (in the direction opposite to the train's motion)

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

(c) F = 1010.62 N = 1.01 KN

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²) / 2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [ (0 m/s) ² - (0.35 m/s) ²] / (2) (2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg) (0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

where,

k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k (2 m)

k = (3368.75 N) / (2 m)

k = 1684.38 N/m = 1.684 KN/m

(c)

Applying Hooke's Law with:

Δx = 0.6 m

F = (1684.38 N/m) (0.6 m)

F = 1010.62 N = 1.01 KN

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg) a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²) / 2s

where,

a = deceleration = - 0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [ (0 m/s) ² - Vi²] / (2) (0.6 m)

Vi = √ (0.0092 m/s²) (1.2 m)

Vi = 0.105 m/s