Calculate the

(i) component of the weight of the cyclist and bicycle parallel to the slope.

(ii) normal reaction force on the bicycle from the slope.

At the bottom of the slope the cyclist has a speed of 5.5ms^–1. The cyclist stops pedalling and applies the brakes which provide an additional decelerating force of 250 N. Determine the distance taken for the cyclist to stop. Assume air resistance is negligible and that there are no other frictional forces.

(i) (weight) = 85 × 9.81(=834N); (if 850 (N) seen, award this mark)
component =(834×sin19=)271 (N);
Allow use of g=10ms^–2. Answer is 277 (N).

(ii) component=(834×cos19=) 788 (N);
Allow use of g=10ms ^–2. Answer is 804 (N).
Allow a bald correct answer.
Do not award ECF if cos used in (a)(i) and sin used in (a)(ii).

total decelerating force =271+250(=521N);
acceleration \( = \left(  -  \right)\frac{{521}}{{85}}\left( { =  - 6.13{\rm{m}}{{\rm{s}}^{ - 2}}} \right)\);
\(s = \frac{{{v^2} - {u^2}}}{{2a}}\);
2.47 (m);} (signs must be consistent for this mark, ie: if acceleration assumed positive, look for negative distance)
Allow use of g=10. Answers are 527 N, 6.2ms^–2, 2.44 m.

or

total decelerating force =271+250(=521N) ;
initial kinetic energy \( = \frac{1}{2}m{v^2} = 1290{\rm{ J}}\)
\({\rm{distance = }}\frac{{{\rm{energy lost}}}}{{{\rm{force}}}} = \frac{{1290}}{{521}}\)
2.47 (m);