Answer:
(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
