Question

A 62-kg bicyclist rides his 10.3-kg bicycle with a speed of 13.5 m/s. Assuming contant acceleration, (a) How much work must be done by the brakes to bring the bike and rider to a stop? kJ (b) How far does the bicycle travel if it takes 4.7 s to come to rest? m (c) What is the magnitude of the braking force? N

Answer 1

Work = energy = force * distance

The energy of a moving bike + biker is in the form of kinetic energy.

Kinetic Energy = (1/2) m v^2

m = mass = mass of bike + mass of biker
m = 10.3 + 62
m = 72.3kgs

v = 13.5m/s

Kinetic energy = 0.5 * 72.3 * 13.5^2
Work done = 6588.3375 J

or 6.58 KJ

Find the distance travelled by an accelerating body:
We know that final velocity (v) = initial velocity (u) + acceleration (a) * time (t)
v = u + at
Final velocity = 0 - we've stopped.
Initial velocity = 13.5 m/s
Time = 4.7 seconds
0 = 13.5 + 4.7a
a = -2.872 m/s^2

Now we know the acceleration, we can find the distance we went in those 4.7 seconds.

Distance (s) = ut + (1/2)at^2
s = 13.5 *4.7 + (1/2) (-2.872) (4.7^2)
s = 63.45 - 31.721
s = 31.729 m

Work = Force * distance
6588.3375 = F * 31.729
F = 6588.3375 / 31.729
F = 207.644 N

verify

The other way to get force is from the equation
Force (F) = mass (m) * acceleration (a)
F = ma
F = 72.3* -2.872 = -207. 644 N

Magnitude of Force = 207. 644 N