Physics Help.A 100 kg physics teacher on his 6 kg bike reaches the top of a hill...

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...moving at 4 m/s. He decides? A 100 kg physics teacher on his 6 kg bike reaches the top of a hill moving at 4 m/s. He decides to take a break and coasts down the other side of the hill.

a) If the hill is 120 m long at a 2o degree incline how fast will he be going at the bottom of the hill?
b) What will be the normal force acting on the teacher and his bike as he costs down the hill?

Thanks for your help.
 
a) sum of forces = mass x acceleration

100kg + 6kg= mass
mgsin20 = acceleration

mgsin20 = 948m/s

v= square root of (vo^2 + 2ad)

v= sqrt (4^2 + (2x948x120))

v= 477m/s

though that seems huge its also because 100kg is one fat teacher

b) fn = y component of gravity

fn = mgcos(20)

423N
 
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