A force of 100 N is applied on a block of mass 3 kg as shown in the figure. The coefficient of friction between the surface and the block is μ=(1/√3) The frictional force acting on the block is

Question

A force of 100 N is applied on a block of mass 3 kg as shown in the figure. The coefficient of friction between the surface and the block is μ=(1/√3) The frictional force acting on the block is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/18777fc7abc25248cd1d63fb3c2579f2-.png" /> (A) 15 N downwards (B) 25 N upwards (C) 20 N downwards (D) 30 N upwards

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/4ffd12bb4487cab5696ce9cdbf6b0b12-.png" /> N'=100 cos 30°=100 ⋅ (√3/2)=50√3N Net driving force =(100×(1/2)-30) = 20 N (upward) f textlimiting (1/√3)(50√3)=50 N ∴ friction = 20 N (downward)

Q. A force of $100 N$ is applied on a block of mass $3 k g$ as shown in the figure. The coefficient of friction between the surface and the block is $μ = \frac{1}{3}$ The frictional force acting on the block is

$15 N$ downwards

$25 N$ upwards

$20 N$ downwards

$30 N$ upwards

$N^{'} = 100 cos 3 0^{\circ} = 100 \cdot \frac{3}{2} = 503 N$

Net driving force $= (100 \times \frac{1}{2} - 30)$

$= 20 N$ (upward)

$f_{limiting} \frac{1}{3} (503) = 50 N$

$∴$ friction $= 20 N$ (downward)

Q. A force of 100N is applied on a block of mass 3kg as shown in the figure. The coefficient of friction between the surface and the block is μ=3​1​ The frictional force acting on the block is

15N downwards

25N upwards

20N downwards

30N upwards