Friction Force – MSQ - Physics MCQ

Let the value of a be increased from zero. As long as a≤μg, there shall be no relative motion between m₁ or m₂ and platform, that is, m₁ and m₂ shall move with accelration a.
As a>μg the acceleration of m₁ and m₂ shall become μgμg each. Hence at all instants the velocity of m₁ and m₂ shall be same The spring shall always remaing in nature length.

The friction force on block A is

The net force on block B in vertical direction is zero

⇒ Normal reaction exerted by ground on block B is 110 N.
frictional force exerted by ground on block B is zero

The correct answers are: The normal reaction exerted by the ground on the block B is 110N, The frictional force exerted by the ground on the block B is zero

The correct answers are: F₁(max) > F₂ (max), F₁(max) : F₂ (max) = 5 : 3

Case-I : μ₁ = 0.5, μ₂ = 0.3
Along the incline, acceleration of 5kg block will be less than acceleration of 3kg block provided they move alone on the incline. The reason is greater friction coefficient of 5kg block, as acceleration along the incline is g sin θ - μg cos θ
Hence, both blocks will move together. In this case FBDs of both are shown.

For 5 kg block

Case II :  acceleration of 5 kg block is more than acceleration of 3 kg. Hence, no normal force will act between them.

The correct answers are: if µ₁ = 0.5 and µ₂ = 0.3 then 5kg block exerts 3N force on the 3kg block, if µ₁ = 0.3 and µ₂ = 0.5 then 5kg block exerts no force on the 3kg block

Block is moving upwards due to friction

Contact force is the result of N and f_r


The correct answers are: Force of friction on block is 6N  upwards, Contact force between the block and belt is 10.5N

FBD in reference frame of the lift


The correct answers are: 

FBD

Net force without friction on system is ‘7N’ in right side so first maximum friction will come on 3 kg block.
So, F₁ = 1 N, F₂ = 6N, T = 2 N.
The correct answers are: f₁ = 1N , f₂ = 6N, T = 2N

Since block M₁ is attached to the string so it will remain stationary. M₂ has tendency to move downward.
So, friction on M₂ will act in upward direction.
N is the force by which the both block are pressed i.e. N = Tcos 30°.

for block M₁

T = 306.75N
T ≈ 306 N
(a = 0 for M₁ since attach to string)
for block M₂

a = 4.67ms^–2
The correct answers are: Tension is 306 N, a_M2 = 4.7 m/s²

But mg = w


The correct answers are: