A cricket is rolled on ice with a velocity of 6 m / s and comes to rest after traveling 10 m. Find the coefficient of friction. ( g =9.8 m / s 2)

Question

A cricket is rolled on ice with a velocity of 6 m / s and comes to rest after traveling 10 m. Find the coefficient of friction. ( g =9.8 m / s 2) (A) -1 (B) -0.18 (C) -0.17 (D) -2

Tardigrade Admin · Accepted Answer

v 2- u 2=2 as a =((02)-(6)2/2 × 10)=(-36/20)=-1.8 a =μ g ⇒ μ=( a / g )=(-1.8/9.8) ⇒ μ=-0.18 (-ve sign indicates that friction force acts in opposite direction.)

Q. A cricket is rolled on ice with a velocity of $6 m / s$ and comes to rest after traveling $10 m$ . Find the coefficient of friction. $(g = 9.8 m / s^{2})$

-1

-0.18

-0.17

-2

$v^{2} - u^{2} = 2 a s$
$a = \frac{( 0 ^{2} ) - ( 6 ) ^{2}}{2 \times 10} = \frac{- 36}{20} = - 1.8$
$a = μg$
$\Rightarrow μ = \frac{a}{g} = \frac{- 1.8}{9.8}$
$\Rightarrow μ = - 0.18$
(-ve sign indicates that friction force acts in opposite direction.)

Q. A cricket is rolled on ice with a velocity of 6m/s and comes to rest after traveling 10m. Find the coefficient of friction. (g=9.8m/s2)

-1

-0.18

-0.17

-2

v2−u2=2as a=2×10(02)−(6)2​=20−36​=−1.8 a=μg ⇒μ=ga​=9.8−1.8​ ⇒μ=−0.18 (-ve sign indicates that friction force acts in opposite direction.)

Q. A cricket is rolled on ice with a velocity of $6 m / s$ and comes to rest after traveling $10 m$ . Find the coefficient of friction. $(g = 9.8 m / s^{2})$

$v^{2} - u^{2} = 2 a s$
$a = \frac{( 0 ^{2} ) - ( 6 ) ^{2}}{2 \times 10} = \frac{- 36}{20} = - 1.8$
$a = μg$
$\Rightarrow μ = \frac{a}{g} = \frac{- 1.8}{9.8}$
$\Rightarrow μ = - 0.18$
(-ve sign indicates that friction force acts in opposite direction.)