Question
A train has a total mass of \(7.5 × 10^5\) kg.
(a) The train accelerates from rest at a constant rate along a straight, horizontal track.
It reaches a speed of 24 m / s in 60 s.
Calculate
(i) the train’s acceleration,
acceleration = …………………………..
(ii) the resultant force acting on the train.
force =
(b) The train now travels with a constant speed of 24 m / s along a straight, horizontal track.
The total force opposing the motion due to friction and air resistance is \(7.2 × 10^4\) N.
(i) By considering the work done by the train’s engine in 1.0 s, calculate its output
power.
power = …………………
(ii) The train begins to travel up a slope.
Explain why the power of the train’s engine must be increased to maintain the
speed of 24 m / s.
Answer/Explanation
Answer:
(a) (i) v = u + at OR (a =) (v – u)/t OR 24 = a × 60 OR 24/60
0.4(0)m/\(s^2\)
(ii) (F =) ma OR \(7.5 × 10^5 × 0.40\)
300 000N OR 300kN
(b) (i) in words or symbols (P =) W/t OR F x d/t OR Fv
OR \(7.2 × 10^4 × 24 / 1 \)OR OR \(7.2 × 10^4 × 24\)
(ii) gravitational/potential energy of train has to be increased
OR force acts down the slope/backward force acts (on train)
(for the same distance moved) more work done has to be done OR energy
has to be provided (by the engine)
in the same time (so needs more power)