# 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 ?

2
by abha970
mass of the cycle + cyclist is 70 kg ?
yes

2015-02-22T13:43:11+05:30

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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.

2015-02-22T13:47:23+05:30
given that ,
m = 70 kg ( mass of the cycle + cyclist )
u = 18 km/h ( initial velocity )
=
v = 9 km/h
= 9 * 1000 /3600 = 2.5 m/s
W =  ?
W1 = ? ( initial work done)
W2= ? ( final work done )

W =
=
= -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 )