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Standard XII

Physics

Net Force

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 net contact force will be

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 $= 12 N$
$∴ f_{m a x} = μ N = 12 \times 0.6 = 7.2 N$
Since weight $W < f_{m a x}$ 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 net contact force will be

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

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 net contact force will be

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