Method 1: Using F=ma and v²=u²+2as
1. Calculate deceleration: a = F / m = 270 000 N / 240 000 kg = 1.125 m/s².
2. Use equation of motion: v² = u² + 2as.
3. Identify variables: u = 60 m/s, v = 0 m/s, a = -1.125 m/s².
4. Rearrange for distance (s): s = (v² - u²) / 2a.
5. Substitute: s = (0² - 60²) / (2 × -1.125) = -3600 / -2.25.
6. Calculate: s = 1600 m.
Method 2: Using Work-Energy Theorem
1. Initial kinetic energy: KE = ½mv² = ½ × 240 000 × 60² = 432 000 000 J.
2. Work done by braking force = Force × distance (W = Fd).
3. To stop the train, work done must equal the initial KE.
4. Fd = 432 000 000 J.
5. Rearrange for distance (d): d = 432 000 000 / F.
6. Substitute: d = 432 000 000 / 270 000 = 1600 m.
This problem can be solved by combining equations for force and motion, or by using the concept of work and energy.
**Method 1: Using F=ma and Equations of Motion**
1. **Calculate deceleration (a):**
Use Newton's Second Law, F = ma. Rearrange to find a:
a = F / m
a = 270 000 N / 240 000 kg
a = 1.125 m/s²
Since this is a braking force, the acceleration is negative: a = -1.125 m/s².
2. **Calculate distance (s):**
Use the equation of motion that relates initial speed (u), final speed (v), acceleration (a), and distance (s):
v² = u² + 2as
We know:
u = 60 m/s (initial speed)
v = 0 m/s (the train stops)
a = -1.125 m/s²
Rearrange to solve for s:
s = (v² - u²) / 2a
s = (0² - 60²) / (2 × -1.125)
s = (-3600) / (-2.25)
s = 1600 m
**Method 2: Using Work-Energy Theorem**
1. **Calculate initial Kinetic Energy (KE):**
The work done by the brakes must equal the initial kinetic energy of the train to bring it to a stop.
KE = ½mv²
KE = 0.5 × 240 000 kg × (60 m/s)²
KE = 120 000 × 3600
KE = 432,000,000 J
2. **Calculate distance (d):**
Work Done (W) = Force (F) × distance (d)
W = KE
Fd = 432,000,000 J
d = 432,000,000 J / 270 000 N
d = 1600 m
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