Question

please asnwer it correctly

Answer 1

Answer:

Let a be the acceleration of block when sliding and distance traveled by the block A from top of wedge to bottom of wedge.

i.e d=lsecα

Friction force, f=kN=kmgcosα

Therefore, mgsinα−kmgcosα=ma........(1)

a=gsinα−kgcosα

Now, from kinematical equation: (d=ut+

2

1

at

2

)

lsecα=0+(

2

1

)at

2

t=

(sinα−kcosα)

2lsecα

g

(using equation (1))

=

(

2

sin2α

−kcos

2

α)

2l

g

......(2)

For t

min

,

dα

d(

2

sin2α

−kcos

2

α)

=0

i.e.

2

2cos2α

+2kcosαsinα=0

cos2α+ksin2α=0

tan2α=−

k

1

⟹α=49

∘

Putting the values of α, k and l in equation (2) we get t

min

=1 s