A force of 100 N is applied on a block of mass 3 kg as shown below- The coefficient of friction between wall and the block is 1-4- The friction force acting on the block isA- 20 N upwardsB- 25 N upwardsC- 15 N downwardsD- 20 N downwards

The correct option is D 20 N downwards f_max=μ N= ( 1/4 ) ( 100√(3)/2 ) =25√(3)/2 Newton Since net force (excluding friction) acting on the block is 20 N u ...

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Byju's Answer

Standard XII

Physics

Calculating Friction Using Coeffcients

A force of 10...

Question

A force of 100 N is applied on a block of mass 3kg as shown below. The coefficient of friction between wall and the block is $\frac{1}{4}$ . The friction force acting on the block is

20 N upwards

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20 N downwards

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25 N upwards

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15 N downwards

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Solution

The correct option is B
20 N downwards

$f_{m a x} = μ N = (\frac{1}{4}) (\frac{100 \sqrt{3}}{2})$

$= \frac{25 \sqrt{3}}{2}$ Newton

Since net force (excluding friction) acting on the block is 20 N upwards therefore f=20N downwards.

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A force of 100 N is applied on a block of mass 3kg as shown below. The coefficient of friction between wall and the block is 14. The friction force acting on the block is

The correct option is B 20 N downwards fmax=μN=(14)(100√32) =25√32 Newton Since net force (excluding friction) acting on the block is 20 N upwards therefore f=20N downwards.

A force of 100 N is applied on a block of mass 3kg as shown below. The coefficient of friction between wall and the block is $\frac{1}{4}$ . The friction force acting on the block is

The correct option is B
20 N downwards

$f_{m a x} = μ N = (\frac{1}{4}) (\frac{100 \sqrt{3}}{2})$

$= \frac{25 \sqrt{3}}{2}$ Newton

Since net force (excluding friction) acting on the block is 20 N upwards therefore f=20N downwards.