Question

the speed of bicycle decreased from 18 km/h to 9km/h . if total mass of th cyclist with bicycle is 70 kg . calculate the work done is overcoming the friction offered by the brakes and the road . does the K.E.of the system also reduce to half its initial value ?

Answer 1

m = 70 kg
v = 9 kmph = 9 * 5/18 = 2.5 m/s
u = 18 kmph = 18 * 5/18 = 5 m/s

Work done  = change in kinetic energy
              = initial KE - final KE
              = 1/2 m u² - 1/2 m v²
              = 1/2 * 70 * 5² - 1/2 * 70 * 2.5²
            = 875 J - 218.75 J  = 656.25  J

 The speed has come down to half its value.  But the Kinetic energy becomes 218.75/875 = 1/4 th of inital KE.

Answer 2

given that ,
 m = 70 kg ( mass of the cycle + cyclist )
 u = 18 km/h ( initial velocity )
    = $\frac{18 * 1000 }{3600} = 5m/s$
 v = 9 km/h
  = 9 * 1000 /3600 = 2.5 m/s
 W =  ? 
 W1 = ? ( initial work done)
 W2= ? ( final work done )

 W = $\frac{1}{2} mv^2 - \frac{1}{2} mu^2$
     = [tex] \frac{1}{2} *70 * 2.5^2 - \frac{1}{2} * 70 *5^2 [/tex]
     = -656.25 Joules 
W1 = over coming force of friction 
therefore , W1 = 656.25 J

W2 = 1/2*70*5^2
     = 875 joules 
    
now,
 W1 / W2 = 656.25 / 875 = 3/4        ( THE K.E. is reduced to 3/4 th of its initial value )