Question

A bicyclist on level ground moving with 13.4 m/s .What is the cyclist’s and bicycle’s mass if the increase in internal energy is 5836 J?

1- 65.0 kg

2- 100 kg

3- 150 kg

4- 200kg

Answer 1

Answer:-

Cyclist rides the bicycle at speed, v=14√3m/s.

radius of circular road , r=20√3m.

Let inclination angle with vertical by cyclist is A.

a normal reaction acts between floor and bicycle.

then at equilibrium,

NsinA=rmv²_____(i)

NcosA=mg_____(ii)

dividing equations (i) by (ii),

tanA=rgv2​

now put v=14√3,r=20√3 and g=9.8m/s

so , tanA=(20√3×9.8)(14√3)2​

=196√3196×3​

=196196√3​

=√3

hence,  tanA=√3=tan60°

so, A=60°

Hope it helps..!!

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