Answer:
(i) 20 m/s
(ii) 0.00625 m/s²
(iii) 25 m/s
Explanation:
(i) The train moves at constant speed when it has no acceleration, meaning when the sum of the forces is zero.
There are two forces acting on the train. Â The force of the engine pushing the train forward, and the resistance.
∑F = ma
F − R = 0
F = R
The resistance is:
50 N/tonne × (100 tonne + 25 tonne) = 6250 N
Therefore, the force of the engine is 6250 N. Â The power of the engine is 125 kW. Â Power is work per time, and work is force times distance, so:
P = W / t
P = Fd / t
P = Fv
125000 Nm/s = (6250 N) v
v = 20 m/s
(ii) The truck becomes uncoupled. Â The engine is still pushing with 6250 N of force, but the resistance has changed to:
50 N/ton × 100 tonne = 5000 N
∑F = ma
F − R = ma
6250 N − 5000 N = (100 tonne × 2000 kg/tonne) a
a = 0.00625 m/s²
(iii) When the force of the engine matches the new resistance force:
P = Fv
125000 Nm/s = (5000 N) v
v = 25 m/s