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Laws of Motion question

2003 · Shift 0 · Q153
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Laws of Motion question

2003 · Shift 0 · Q153

JEE MainPhysicsLaws of MotionMCQ+4 / −1
A horizontal force of 10N10N10N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.20.20.2. The weight of the block is AIEEE 2003 Physics - Laws of Motion Question 121 English
  1. A
    20N20N20N
  2. B
    50N50N50N
  3. C
    100N100N100N
  4. D
    2N2N2N
View written solutionFree

Correct answer: D

  1. Identify the forces on the block

A block is pressed horizontally against a vertical wall by a force of 10 N10\,\text{N}10N.

  • Horizontal applied force: 10 N10\,\text{N}10N toward the wall
  • Normal reaction from wall: NNN
  • Friction force: acts vertically upward to prevent the block from sliding down
  • Weight of block: WWW downward
  1. Find the normal reaction

Since the block is stationary in the horizontal direction,

N=10 NN = 10\,\text{N}N=10N

  1. Use limiting friction condition

The force is said to be just sufficient to hold the block stationary, so friction is at its maximum value:

fmax⁡=μNf_{\max} = \mu Nfmax​=μN

Given:

μ=0.2,N=10 N\mu = 0.2, \quad N = 10\,\text{N}μ=0.2,N=10N

So,

fmax⁡=0.2×10=2 Nf_{\max} = 0.2 \times 10 = 2\,\text{N}fmax​=0.2×10=2N

  1. Balance vertical forces

For the block to remain at rest,

f=Wf = Wf=W

At the limiting condition,

W=2 NW = 2\,\text{N}W=2N

  1. Conclusion

The weight of the block is

2 N\boxed{2\,\text{N}}2N​

So the correct option is D.

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