Question

a 65kg bicyclists rides his 8.8kg bicycle with a speed of 14m/s...(a) how much work must be done by the brakes to bring the bike and rider to a stop? (b) how far does the bicycle travel if it takes 4sec to come to rest? (c) what is the magnitude of the braking force?

Answer 1

Explanation:

It is given that,

Mass of bicyclists, m₁ = 65 kg

Mass of bicycle, m₂ = 8.8 kg

Speed, u = 14 m/s

(a) For calculating work done by the brakes to bring the bike and the rider to a stop. It can be calculated by finding the change in kinetic energy as per work- energy theorem.

$W=\dfrac{1}{2}m(v^2-u^2)$

m = m₁ + m₂ = 73.8 kg

$W=\dfrac{1}{2}\times 73.8\ kg\times (0-(14\ m/s)^2)$    

W = -7232.4 Joules

(b) From first equation of motion :

v = u + at

0 = 14 + a × 4

a = -3.5 m/s²

Let s is the distance travelled by the bicycle. Now, using third equation of motion as :

$v^2-u^2=2as$

$\dfrac{v^2-u^2}{2a}=s$

$\dfrac{0-(14\ m/s)^2}{2\times (-3.5\ m/s^2)}=s$

s = 28 m

The bicycle will move 28 m if it takes 4 seconds to come to rest.

(c) Braking force, F = m a

So, F = 73.8 kg × -3.5 m/s²

F = −258.3 N

Hence, this is the required solution.

Answer 2

(a) The total mass of the bicycle+rider system is

m=65 kg+8.8 kg=73.8 kg

And their initial kinetic energy is

$K= \frac{1}{2}mv^2= \frac{1}{2}(73.8 kg)(14 m/s)^2=7232 J$

For the work-energy theorem, the work done by the brakes to stop the bicycle+rider must be equal to the loss of kinetic energy of the system:

$W=K_f -K_i$

And so, since the final kinetic energy Kf is zero (because the  final velocity is zero), then we have

$W=-K_i = -7232 J$

Where the negative sign is due to the fact that the work done by the brakes is against the motion of the bicycle.

(b) First we can calculate the acceleration of the bicyle; since the final speed is zero, we have

$a= \frac{v_f-v_i}{t} = \frac{-14 m/s}{4 s}=-3.5 m/s^2$

where the negative sign means it is a deceleration.

Then we can find the distance covered by using the relationship:

$2aS = v_f^2 - v_i^2$

So we have

$S= \frac{-v_i^2}{2a}= \frac{-(14 m/s)^2}{2 (-3.5 m/s^2)}=28 m$

So, the bike and the rider traveled for 28 meters before to stop.

(c) The work done by the brake force is:

$W=Fd$

where F is the brake force and d is the distance covered by the bike+rider during the time of application of the force, so it is d=S=28 m.

Therefore, the brake force is (we can neglect signs because we are only interested in the magnitude of the force):

$F= \frac{W}{d}= \frac{7232 J}{28 m}=258.3 N$