. A particle is projected up along a rough inclined plane of inclination 45∘ with the horizontal. If the coefficient of friction is 0.5, the acceleration is a. 2g b. 22g c. 223g d. 2g

Sure, here are the step by step solutions: Given: θ=45∘ μ=0.5 Acceleration, a=g(sinθ+μcosθ) Substituting the given values: a=g(sin45∘+0.5cos45∘) Since sin45∘=cos45∘=21: a=g(21+0.5⋅21) Simplifying the expression: a=g(21+20.5) a=g(21.5) Multiplying both numerator and denominator by 2 to rationalize the denominator: a=21.5g2 Simplifying the expression: a=223g Therefore, the acceleration is 223g.

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. A particle is projected up along a rough inclined plane of inclination $45^{\circ}$ with the horizontal. If the coefficient of friction is 0.5, the acceleration is a. $\frac{g}{2}$ b. $\frac{g}{2 2}$ c. $\frac{3 g}{2 2}$ d. $\frac{g}{2}$

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Sophia Discussed

. A particle is projected up along a rough inclined plane of inclination

45^{\circ}

with the horizontal. If the coefficient of friction is 0.5, the acceleration is a.

\frac{g}{2}

\frac{g}{2 2}

\frac{3 g}{2 2}

\frac{g}{2}

7 mins ago

Discuss this question LIVE

7 mins ago

Text solutionVerified

Sure, here are the step by step solutions: Given:

θ = 45^{\circ}

μ = 0.5

Acceleration,

a = g (sin θ + μ cos θ)

Substituting the given values:

a = g (sin 45^{\circ} + 0.5 cos 45^{\circ})

Since

sin 45^{\circ} = cos 45^{\circ} = \frac{1}{2}

a = g (\frac{1}{2} + 0.5 \cdot \frac{1}{2})

Simplifying the expression:

a = g (\frac{1}{2} + \frac{0.5}{2})

a = g (\frac{1.5}{2})

Multiplying both numerator and denominator by

2

to rationalize the denominator:

a = \frac{1.5 g 2}{2}

Simplifying the expression:

a = \frac{3 g}{2 2}

Therefore, the acceleration is

\frac{3 g}{2 2}

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Question Text	. A particle is projected up along a rough inclined plane of inclination $45^{\circ}$ with the horizontal. If the coefficient of friction is 0.5, the acceleration is a. $\frac{g}{2}$ b. $\frac{g}{2 2}$ c. $\frac{3 g}{2 2}$ d. $\frac{g}{2}$
Topic	Circular & Rotation Motion
Subject	Physics
Class	Class 12
Answer Type	Text solution:1

. A particle is projected up along a rough inclined plane of inclination $45^{\circ}$ with the horizontal. If the coefficient of friction is 0.5, the acceleration is a. $\frac{g}{2}$ b. $\frac{g}{2 2}$ c. $\frac{3 g}{2 2}$ d. $\frac{g}{2}$

Text solutionVerified

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. A particle is projected up along a rough inclined plane of inclination $45^{\circ}$ with the horizontal. If the coefficient of friction is 0.5, the acceleration is a. $\frac{g}{2}$ b. $\frac{g}{2 2}$ c. $\frac{3 g}{2 2}$ d. $\frac{g}{2}$