A block of mass 0.5 kg has an initial velocity of 10 m s -1 down an inclined plane of angle 30°, the coefficient of friction between the block and the inclined surface is 0.2 The velocity of the block after it travels a distance of 10 m is (Take .g=10 m s -2)

Question

A block of mass 0.5 kg has an initial velocity of 10 m s -1 down an inclined plane of angle 30°, the coefficient of friction between the block and the inclined surface is 0.2 The velocity of the block after it travels a distance of 10 m is (Take .g=10 m s -2) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/5b0ab3dbbd32bc0d0eed1520b5d715f2-.png" /> (A) 17 m s -1 (B) 13 m s -1 (C) 24 m s -1 (D) 8 m s -1

Tardigrade Admin · Accepted Answer

Here, m=0.5 kg , u=10 ms -1, θ=30°, μ=0.2, s=10 m Acceleration down the plane, a=g( sin θ-μ cos θ)=10( sin 30°-0.2 cos 30°)=3.268 text From, v2=u2+2 a s=102+2(3.268) × 10=165.36 or v=√165.36 ≈ 13 m s -1

Q. A block of mass $0.5 k g$ has an initial velocity of $10 m s^{- 1}$ down an inclined plane of angle $3 0^{\circ},$ the coefficient of friction between the block and the inclined surface is $0.2$ The velocity of the block after it travels a distance of $10 m$ is (Take $g = 10 m s^{- 2})$

$17 m s^{- 1}$

$13 m s^{- 1}$

$24 m s^{- 1}$

$8 m s^{- 1}$

Here, $m = 0.5 k g, u = 10 m s^{- 1}, θ = 3 0^{\circ}, μ = 0.2, s = 10 m$
Acceleration down the plane,
$a = g (sin θ - μ cos θ) = 10 (sin 3 0^{\circ} - 0.2 cos 3 0^{\circ}) = 3.268$
$From, v^{2} = u^{2} + 2 a s = 1 0^{2} + 2 (3.268) \times 10 = 165.36$
or $v = 165.36 \approx 13 m s^{- 1}$

Q. A block of mass 0.5kg has an initial velocity of 10ms−1 down an inclined plane of angle 30∘, the coefficient of friction between the block and the inclined surface is 0.2 The velocity of the block after it travels a distance of 10m is (Take g=10ms−2)

17ms−1

13ms−1

24ms−1

8ms−1