Question

hva block to slide down a rough inclined plane at 45° to the horizontal is twice as compared
The time taken by a block to slide down a rough inclined
the time required by the block to slide down smooth frictionless plane inclined at 45° to the horizont
coefficient of kinetic friction between the block and rough plane is

Answer 1

Answer:

Here I have taken n= √2times from the question it is clear that T(rough) =√2 T(smooth) t1= √2 t2 From the fbd (For rough surface) N = mgcos45...........(1). And. f = uN.........(2) From Newton law Net F = ma1 mg(sin45)- u mg(cos 45) =ma1 Solving...... a1 =( 1-u)g/√2 (For smooth surface) mgsin45 = ma2 a2 = g/√2 Now from the common sense we can assume that both the block have to cover the same distance now using the equation of kinematics S = ut + 1/2 at^2. Where u = 0 m/s S1 = 1/2*(1-u)g/√2*(√2t)^2. (Where * means multiply) S2= g/√2 *(t)^2 Now S1 =S2 U=1/2(this is not the answer to this value but u can solve this way And answer is 0.75)

Answer 2

Answer:

$Acceleration along the downward directioon of wedgeα 1​ m(mgsin45 ∘ −μmg cos45 ∘ )​ = 2​ (1−μ)g​ Let total time taken is t,S=o×t+0.5(μ)t 2 when wedge is friction lessthen accelerationa 2​ = 2​ g​ 8=o× 2t​ +.5(a 2​ )×( 2t​ ) 2 Equate both value of Sμ= 43​ Hence (B) is correct answer$

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