A block starts moving up a fixed inclined plane of inclination text60° with a velocity of 20 ms- 1 and stops after text2 s . The approximate value of the coefficient of friction is [ g=10 ms- 2 ]

Question

A block starts moving up a fixed inclined plane of inclination text60° with a velocity of 20 ms- 1 and stops after text2 s . The approximate value of the coefficient of friction is [ g=10 ms- 2 ] (A)  text3 (B)  text3 text.3 (C)  text0 text.27 (D)  text0 text.33

Tardigrade Admin · Accepted Answer

Retardation of the block is a=g( textsin θ + μ cos θ )=5(√3 + μ ) v=u-at=0⇒ a=(u/t)⇒ 5(√3 + μ )=(20/2)⇒ μ =0.27

Q. A block starts moving up a fixed inclined plane of inclination $60^{^{\circ}}$ with a velocity of $20 m s^{- 1}$ and stops after $2 s$ . The approximate value of the coefficient of friction is [ $g = 10 m s^{- 2}$ ]

$3$

$3 .3$

$0 .27$

$0 .33$

Retardation of the block is
$a = g (sin θ + μ cos θ) = 5 (3 + μ)$
$v = u - a t = 0 \Rightarrow a = \frac{u}{t} \Rightarrow 5 (3 + μ) = \frac{20}{2} \Rightarrow μ = 0.27$

Q. A block starts moving up a fixed inclined plane of inclination 60∘ with a velocity of 20ms−1 and stops after 2 s . The approximate value of the coefficient of friction is [ g=10ms−2 ]

3

3.3

0.27

0.33

Retardation of the block is a=g(sinθ+μcosθ)=5(3​+μ) v=u−at=0⇒a=tu​⇒5(3​+μ)=220​⇒μ=0.27

Q. A block starts moving up a fixed inclined plane of inclination $60^{^{\circ}}$ with a velocity of $20 m s^{- 1}$ and stops after $2 s$ . The approximate value of the coefficient of friction is [ $g = 10 m s^{- 2}$ ]

$3$

$3 .3$

$0 .27$

$0 .33$

Retardation of the block is
$a = g (sin θ + μ cos θ) = 5 (3 + μ)$
$v = u - a t = 0 \Rightarrow a = \frac{u}{t} \Rightarrow 5 (3 + μ) = \frac{20}{2} \Rightarrow μ = 0.27$