The force applied on a body on a rough surface of coefficient of friction $μ$ produces an acceleration $a$ . If coefficient of friction between the body and the surface is reduced to $μ / 3$ , the acceleration increases by $2$ units. The value of $μ$ is:

\frac{2}{3} g

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\frac{3}{2} g

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\frac{3}{g}

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\frac{1}{g}

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Solution

The correct option is D $\frac{3}{g}$
Case 1:
$a = μ g$
Case 2:
$a + 2 = \frac{μ}{3} g$
Dividing both we get:
$\frac{a}{a + 2} = 3$
$∴ a = - 3$
From above equation we have
$μ = - \frac{3}{g}$
Considering only magnitude:
$∴ μ = \frac{3}{g}$

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The force applied on a body on a rough surface of coefficient of friction μ produces an acceleration a. If coefficient of friction between the body and the surface is reduced to μ/3, the acceleration increases by 2 units. The value of μ is:

The correct option is D 3gCase 1:a=μgCase 2:a+2=μ3gDividing both we get:aa+2=3∴a=−3From above equation we haveμ=−3gConsidering only magnitude:∴μ=3g

The force applied on a body on a rough surface of coefficient of friction $μ$ produces an acceleration $a$ . If coefficient of friction between the body and the surface is reduced to $μ / 3$ , the acceleration increases by $2$ units. The value of $μ$ is:

The correct option is D $\frac{3}{g}$
Case 1:
$a = μ g$
Case 2:
$a + 2 = \frac{μ}{3} g$
Dividing both we get:
$\frac{a}{a + 2} = 3$
$∴ a = - 3$
From above equation we have
$μ = - \frac{3}{g}$
Considering only magnitude:
$∴ μ = \frac{3}{g}$