The Student Room Group
Reply 1
Using the conservation of energy for the cyclist:
mgs*cos (arcsin 1/20) = F*s
F = mg cos(arcsin 1/20) = 80 *9.81*cos(acrsin 1/20)
= 783.8 N
Is that your answer?
Reply 2
Opps, I am confused

The right thing is:
Using the conservation of energy for the cyclist: the change of energy = work done by friction at the end. Gravitational potential energy = work done by friction
mgs*sin(alpha) = F*s with sin (alpha) = 1/20
F = mg*1/20 = 80 *9.81*1/20
= 39.24 N

I am very sorry about that.
Reply 3
At the top of the hill:
PE = mgh = 80g(80*sin[arcsin(1/20)]) = 320g
KE = 0
At the bottom of the hill:
PE = 0
KE = 0.5mv² = 40v²

Conservation of energy gives:
KE gained + PE lost = work done against resistance
40v² - 320g = Rd = 80R (1)

Now conservation of energy again, this time between the points at the bottom of the hill and at the end of the path:
KE lost = work done against resistance
- 40v² = 80R (2)

(1) = (2):
80v² = 320g
40v² = 160g (3)

Substituting (3) in (1):
160g - 320g = 80R
R = -2g

So the resistance against motion is 2g N.
Reply 4
Hi, I think I have the correct solution to your problem.
Let A be the point at the top of the hill.
Let B be the point at the base of the hill.
Let C be the final point where the cyclist stops.

At A, PE=80*g*4=320g J, since sin(theta)=height/80=1/20, so height=4.
KE=0.
TOTAL ENERGY=320g J.

At B, PE=0
KE=1/2*80*v^2
TOTAL ENERGY= 40v^2.

The work done in going from A to B is Fd=80F N
Hence 320g=40v^2+80F. (1)

NEXT
The TOTAL ENERGY at B is 40v^2 J,(as before)

At C, the PE=0
KE=0

The Work done in going from B to C is Fd=80F N.
Hence 40v^2=0+80F

Substituting this into expression (1), gives
320g=80F+80F
Hence the resistive force F=2g N.
Hope this helps.
Reply 5
Energy at initial point - Energy at final point = Resistance ( work external )
energy at initial point = P.E. only
energy at final pont = 0 ( P.E = 0 ( taking horizontal road as reference, at rest ==> K.E. = 0 )

so:
Resistance = P.E. = m . g . h
h = 80 . sin @
where @ = incline angle
so Resistance = 80 . g . 4 = 320.g
resitance = F.d
d: distance moved
320g = F. 160
F = 32g/16 N

glad to be corrected!
Reply 6
thnx