Is friction the vital part to gain maximum acceleration?

Question:A cyclist and her bicycle have mass 75 kg.She is riding on a horizontal road,and positions herself so that 60% of the normal contact acts on the back wheel and 40% on the front wheel.The coefficent of friction between tyres and the raod is 0.8.What is the greatest acceleration she can hope to acheive?
please make clear this concept that how we can gain maximum acceleration and how it depends on Friction

Since it is the back wheel of the bicycle that provides the force to move (i.e. the front wheel just rotates freely), our analysis need only consider the contact point of the back wheel with the road.
There are three forces acting on the bicycle:
The road is pushing up against the wheel with a normal force N, the road is pushing forward on the wheel with the force of static friction F, and gravity is pulling the bicycle/wheel/rider downwards with weight force W.
We then have:
W = 0.6*M*g
F_static < mu*N
M = 75 kg
mu = 0.8
g = 9.81 m/s/s

The 0.6 appears in the equation for W since only 60% of the total weight is pushing down on the back wheel (i.e. the wheel we are considering).

Newton's 2nd Law in the horizontal direction tells us that:
M*a = F_static
since F_static is the only horizontal force.

Newton's 2nd Law in the vertical direction tells us that:
0 = M*A = N - W
since the sum of the forces must equal mass M times a vertical acceleration of A=0 (i.e. the bicycle is not accelerating upwards or downwards).

Thus we have:
0 = N - W
W = 0.6*M*g
F_static < mu*N
M*a = F_static
M = 75 kg
mu = 0.8
g = 9.81 m/s/s

from which we can write:
M*a = F_static < mu*N = mu*W = mu*0.6*M*g

and so:
M*a < mu*0.6*M*g

dividing both sides by M yields:
a < mu*0.6*g
a < 0.8*0.6*9.81
a < 4.7088 m/s/s

Notice that the equation for *static* friction that I used was:
F_static < mu*N
and NOT:
F_static = mu*N

The second equation is a special case in which one is just on the verge of beginning to slip, and in this special case it should really be written:
F_static_maximum = mu*N

Once slipping starts then we have to use another equation that looks very similar:
F_kinetic = mu_k*N

In the slipping case (kinetic friction) one always has an equal sign, not a greater than sign. One also has a different coefficient of friction when slipping, and so mu_k is not equal to mu. In fact, it is usually the case that:
mu > mu_k
and so one can stop a car faster on ice if one can avoid slipping and get F_static up as large as possible without reaching F_static_maximum. This is one of the two main benefits of antilock breaks. I leave it for you to think about what the other reason is -- the more important of the two reasons.

So the three valid equations for friction are:
F_static < mu*N
F_static_maximum = mu*N
F_kinetic = mu_k*N
and it is usually the case that:
mu > mu_k

I hope that this lesson on friction helps you out.

The best she can hope for would be 60% of .8, or .48 g's. That assumes she can produce enough power.

acceleration will be maximum if force is maximum. (f=ma)
The only force that can make her move forward is frictional force.
Frictional force is dependent on the normal reaction which in turn depends on the mass at the point of contact.

at one tyre f=umg= 0.8x (60% of 75) x 10 = 360N
at second tyre, f=umg = 0.8 x (40% of 75) x 10 = 240N

net frictional force to move her fwd= 600 N

now, mass = 75
so F=ma
a= F/m = 600/75 = 8 m/s^2----------------------ans

mechanics is easy, u just need to know the net force in a direction and calculate stuff accordingly. if theres force, there will be motion (i mean NET force here)