On the diagram above, draw labelled arrows to represent the vertical forces that act on the railway engine.

Explain, with reference to Newton’s laws of motion, why the velocity of the railway engine is constant.

The constant horizontal velocity of the railway engine is 16 ms^–1. A total horizontal resistive force of 76 kN acts on the railway engine.

Calculate the useful power output of the railway engine.

The power driving the railway engine is switched off. The railway engine stops, from its speed of 16 ms^–1, without braking in a distance of 1.1 km. A student hypothesizes that the horizontal resistive force is constant.

Based on this hypothesis, calculate the mass of the railway engine.

Another hypothesis is that the horizontal force in (c) consists of two components. One component is a constant frictional force of 19 kN. The other component is a resistive force F that varies with speed v where F is proportional to v³.

(i) State the value of the magnitude of F when the railway engine is travelling at 16 ms^–1.

(ii) Determine the total horizontal resistive force when the railway engine is travelling at 8.0 ms^–1.

On its journey, the railway engine now travels around a curved track at constant speed. Explain whether or not the railway engine is accelerating.

The shaded box shows the acceptable range of position for W/mg.
single downward arrow labelled W/weight or mg/gravity force; (do not allow gravity)
two upward arrows labelled reaction/contact forces; (do not allow for only one arrow seen)
arrow positions as shown in diagram;

horizontal forces have resultant of zero; (must describe or imply horizontal force)
valid statement linked to theory (e.g. Newton 1/Newton 2/conservation of momentum)
explaining why zero force results in constant velocity/zero acceleration;

power =16×76000;
1.2 MW;

acceleration\( = \frac{{{{16}^2}}}{{2 \times 1100}}\left( { = 0.116} \right)\);
\(m = \left( {\frac{{7.6 \times {{10}^4}}}{{0.116}} = } \right)6.5 \times {10^5}{\rm{kg}}\);
Award [2] for a bald correct answer.

or

use of Fs=\(\frac{1}{2}m{v^2}\);
\(m = \left( {\frac{{2 \times 7.6 \times {{10}^4} \times 1100}}{{{{16}^2}}} = } \right)6.5 \times {10^5}{\rm{kg}}\);
Award [2] for a bald correct answer.

(i) 57 kN;

(ii) \({F_8} = \frac{{{F_{16}}}}{{{2^3}}}\);
F₈=7.1(kN);
total force =19+7.1(kN);
=26 kN;
Award [4] for a bald correct answer.

or

\(k = \left( {\frac{{57 \times {{10}^3}}}{{{{16}^3}}}} \right) = 13.91\);
F₈=(13.91×8³)=7.1(kN);
total force=19+7.1(kN);
=26 kN;
Award [4] for a bald correct answer.

direction of engine is constantly changing;
velocity is speed + direction / velocity is a vector;
engine is accelerating as velocity is changing;
Award [0] for a bald correct answer.

or

centripetal force required to maintain circular motion;
quotes Newton 1/Newton 2;
so engine is accelerating as a force acts;
Award [0] for a bald correct answer.