As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle 30°, with horizontal. For μ s =0.25, the block will just start to move for the value of F: [Given .g=10 ms -2] is

Question

As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle 30°, with horizontal. For μ s =0.25, the block will just start to move for the value of F: [Given .g=10 ms -2] is (A) 20 N (B) 35.7 N (C) 33.3 N (D) 25.2 N

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/fe04858bf94a05090fc5502bfa7919b1-.png" /> N = Mg - F operatornameSin 30° = mg -( F /2)=100-( F /2)=(200- F /2) F operatornameCos 30°=μ N √3 ( F /2)=0.25 ×((200- F /2)) 4 √3 F =200- F F =(200/4 √3+1)=25.22

Q. As shown in the figure a block of mass $10 k g$ lying on a horizontal surface is pulled by a force $F$ acting at an angle $3 0^{\circ}$ , with horizontal. For $μ_{s} = 0.25$ , the block will just start to move for the value of $F$ : [Given $g = 10 m s^{- 2}]$ is

$20 N$

$35.7 N$

$33.3 N$

$25.2 N$

$N = M g - F Sin 3 0^{\circ}$
$= m g - \frac{F}{2} = 100 - \frac{F}{2} = \frac{200 - F}{2}$
$F Cos 3 0^{\circ} = μ N$
$3 \frac{F}{2} = 0.25 \times (\frac{200 - F}{2})$
$43 F = 200 - F$
$F = \frac{200}{4 3 + 1} = 25.22$

Q. As shown in the figure a block of mass 10kg lying on a horizontal surface is pulled by a force F acting at an angle 30∘, with horizontal. For μs​=0.25, the block will just start to move for the value of F : [Given g=10ms−2] is

20N

35.7N

33.3N

25.2N