IGCSE Physics (0625) 1.7.4 Power Paper 3

Question

(a) On a day with no wind, a fountain in Switzerland propels 30 000 kg of water per minute to a
height of 140 m.
Calculate the power used in raising the water.
power = ………………
(b) The efficiency of the pump which operates the fountain is 70%.
Calculate the power supplied to the pump.
power = ………………………..
(c) On another day, a horizontal wind is blowing. The water does not rise vertically.
Explain why the water still rises to a height of 140 m.

Answer/Explanation

Answer:

(a) Fd OR weight × d OR mgh OR 30000 × 10 × 140 OR \(4.2 × 10^7\) seen anywhere
(P = ) E/ t OR W/ t OR mgh/ t symbols or words
\(4.2 × 10^7 / 60\)
\(7.0 ×10^5\) W/ 700 kW/ 0.7 MW

(b) efficiency = output / input OR \((P_{in} =) 100 × P_{out}\) / efficiency
\((P_{in} =) 100 × 7 × 10^5 / 70\)
\(1.0 × 10^6\) W OR 1 000000 W OR 1.0 MW

(c) (horizontal) wind has no effect on P.E gained/ vertical force on water
OR same upward/ vertical force acts on water
OR force from wind is horizontal

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)

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