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Mechanics 2 question

A cyclist starts from rest and freewheels down a hill inclined at arcsin 1/20 to the horizontal. After travelling 80 meters the road becomes horizontal and the cyclist travels a further 80 meters before coming to rest without using the brakes. The cyclist and the bike have a combined mass of 80kg. Find the resistive force (assumed constant throughout the motion).

Anyone fancy a crack at that badboy? I don't see where I'm going wrong.
Reply 1
Avatar for Ollie901
Ollie901
1
Original post by SN-RAVER
X


before I post my working
Is it 19.6N?
(edited 12 years ago)
Reply 2
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SN-RAVER
OP
Original post by Ollie901

Original post by Ollie901
before I post my working
Is it 19.6N?


it is indeed, i got double that though.. don't know where I'm getting that from..
Reply 3
Avatar for Ollie901
Ollie901
1
Original post by SN-RAVER
it is indeed, i got double that though.. don't know where I'm getting that from..


Alright this is how I did it.

The forces acting on the bike are its weight directly downwardsm the reaction force acting perpendicular to the slope and the resistive force up the slope.

Taking the slope and its perpendicular as the horizontal/vertical plane,
The force acting down the slope is the horizontal componant of the weight - the resistance which I called Fr.

= 80g(1/20) - Fr.

putting this equal to ma,

80a=80g(1/20) - Fr

therefore a = (80g/1600) - (Fr/80)
This acts for 80m.

Using v^2 = u^2 + 2as (u=0, s=80)

V = root(160a) where a is what I said it was above.

Now the bike comes off the slope,
u=root(160a)
v=0
s=80

And the new acceleration is -Fr/80 as the only force acting in the horizontal directions is the resistive force. (Fr = ma)

Using v^2 = u^2 + 2as again

0=160((80g/1600)-(Fr/80)) - 80(2Fr/80)
0=160((80g/1600)-(Fr/80)) - 2Fr
0=8g - 2Fr - 2Fr
Fr = 2g
(edited 12 years ago)

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