Two blocks $A$ and $B$ , each of mass $10 k g$ , are sliding on an incline. They are connected by a string. The coefficients of kinetic friction between block A and incline is $μ_{1} = 0.5$ and that between block B and incline is $μ_{2} = 0.4$ . The tension in string will be (take $g = 10 m / s^{2})$

Zero

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2 N

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8 N

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4 N

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Solution

The correct option is D $4 N$
Free body diagram of A

$T + 60 - 40 = 10 a$
$T + 20 = 10 a$ .... $(1)$
Free body diagram of $B$

$60 - T - 32 = 10 a$
$28 - T = 10 a$ ....... $(2)$
From $(1)$ and $(2)$
$28 - T = T + 20$
$2 T = 8$
$T = 4 N$

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Q. Figure (6−E6) shows two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg and the incline is μ₁, and that between the block of mass 4.0 kg and incline is μ₂. Calculate the acceleration of the 2.0 kg block if (a) μ₁ = 0.20 and μ₂ = 0.30, (b) μ₁ = 0.30 and μ₂= 0.20. Take g = 10 m/s².

Figure

Figure shows two blocks in contact sliding down an inclined surface ofinclination 30 $^{\circ}$ . The friction coefficient between the block of mass 2.0 kg and the incline is $μ$ $_{1}$ , and that between the block of mass 4.0 kg and the incline is $μ$ $_{2}$ . Calculate the acceleration of the 2.0 kg block if (a) $μ$ $_{1}$ = 0.20 and $μ$ $_{2}$ = 0.30, (b) $μ$ $_{1}$ = 0.30 and $μ$ $_{2}$ = 0.20. Take g = 10 m/s $^{2}$ .