(Newton's Law of Motion) A 100 kg motorcycle carrying a 70 kg rider comes to a stop in 40 m from a speed of 50 km/h when its brakes are applied. Find the force exerted by the brakes and the force experienced by the rider.

Question

anonymous · Answer

there are 2 ways of solving this question -
1-1/2mv^2 kinetic engery formula 
just subsitute the values
1/2 100+70x(50)^2
u get energy rite
as u know wordone is in joules so is energy so take kinetic energy as workdone 
workdone =fxdistance
f=wordone/distance
2 second method
find acceleration
v^2=u$$v ^{2}=u ^{2}+2as$$
find acceleation 
f=MA
find the force..

anonymous · Answer

use equation $$v ^{2}-u ^{2}=2as$$

v=0m/s   , u=50km/h= 125/9 m/sec 
 
s=40m you get accl =-25/144m/sec

F=MA 

the force exerted by the brakes = Total mass$$\times$$accl=(100+70)-25/144

force experienced by the rider= 70$$\times$$-25/144

anonymous · Answer

v^2-u^2=2ad$$\rightarrow$$a=25.1m/s62
f=ma\[\rightarrow \f=170*25.1=4267 
f=ma  f=70*25.1=1757 n