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Standard XII

Physics

Static Friction

A block of we...

Question

A block of weight $5 N$ is pushed against a vertical wall by a force $12 N$ . The coefficient of friction between the wall and block is 0.6. The magnitude of the force exerted by the wall on the block is

12 N

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5 N

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7.2 N

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13 N

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Solution

The correct option is D $13 N$
$N =$ applied force $= 12 N$
$∴ f_{max} = μ N = 0.6 \times 12 = 7.2 N$
Since, weight $w < f_{max}$
$∴$ Force of friction $f = 5 N$
$∴$ Net contact force $= \sqrt{N^{2} + f^{2}}$
$= \sqrt{(12)^{2} + (5)^{2}} = 13 N$

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A block of weight 5 N is pushed against a vertical wall by a force 12 N. The coefficient of friction between the wall and block is 0.6. The magnitude of the force exerted by the wall on the block is

The correct option is D 13 NN= applied force =12 N ∴fmax=μN=0.6×12=7.2 N Since, weight w<fmax ∴ Force of friction f=5 N ∴ Net contact force =√N2+f2 =√(12)2+(5)2=13 N

A block of weight $5 N$ is pushed against a vertical wall by a force $12 N$ . The coefficient of friction between the wall and block is 0.6. The magnitude of the force exerted by the wall on the block is

The correct option is D $13 N$
$N =$ applied force $= 12 N$
$∴ f_{max} = μ N = 0.6 \times 12 = 7.2 N$
Since, weight $w < f_{max}$
$∴$ Force of friction $f = 5 N$
$∴$ Net contact force $= \sqrt{N^{2} + f^{2}}$
$= \sqrt{(12)^{2} + (5)^{2}} = 13 N$