F.4 physics....

23 ) a)
v = u + at

0 = 30 + 25a
a = -1.2 ms-2

as the train is travelling at a constant speed ,

0 - frictional force = ma
frictional force = 8x10^6 ( 1.2 )
= -9600000 N

b) kinetic energy changed to workdone against friction

1 / 2 m v^2 = Fs

4 x 10^6 x(30^2 ) = -9600000 x s

s = 375m

24) a ) no net force , so the reading = 823N

b )
mg - R = ma
823 - R= 82.3 x 2
R = 658.4 N

c)
R - mg = ma
R = 164.6 + 823
= 987.6 N

26) a ) a = slope = 6 / 8 = 0.75 ms-2

F = ma
= 56 x 0.75
= 42 N

b) mg sinX - f = 42
560sinX = 182
sinX = 0.325
X = 18.97度

inclination = 18.97度

30 ) 唔明佢問咩