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Q8. A bike is moving with a velocity of 100 km/h after applying breaks it stops in 10 secs. Calculate the force applied by the breaks . Mass of the motorcycle is 200 kg.

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Solution

Dear Student,

$u = 100 k m / h = 100 \times \frac{5}{18} = 27.77 m / s e c t = 10 s e c v = 0 k m / h a s i t s t o p s f i n a l l y v = u - a t 0 = 27.77 - a \times 10 a = 2.77 m / s e c^{2} F o r c e = m \times a = 200 \times 2.77 = 554 N R e g a r d s$

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Solve with steps fast ​Q8. A bike is moving with a velocity of 100 km/h after applying breaks it stops in 10 secs. Calculate the force applied by the breaks . Mass of the motorcycle is 200 kg.

Dear Student, u=100km/h=100×518=27.77m/sect=10secv=0 km/h as it stops finallyv=u-at0=27.77-a×10a=2.77m/sec2Force =m×a =200×2.77=554NRegards

Solve with steps fast

Q8. A bike is moving with a velocity of 100 km/h after applying breaks it stops in 10 secs. Calculate the force applied by the breaks . Mass of the motorcycle is 200 kg.

Dear Student,

$u = 100 k m / h = 100 \times \frac{5}{18} = 27.77 m / s e c t = 10 s e c v = 0 k m / h a s i t s t o p s f i n a l l y v = u - a t 0 = 27.77 - a \times 10 a = 2.77 m / s e c^{2} F o r c e = m \times a = 200 \times 2.77 = 554 N R e g a r d s$