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 ...

You visited us 5 times! Enjoying our articles? Unlock Full Access!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20 N downwards

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

25 N upwards

No worries! We‘ve got your back. Try BYJU‘S free classes today!

15 N downwards

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

Similar questions

Q. A force of

100 N

is applied on a mass of

3 k g

as shown in the figure. The coefficient of friction between the surface and the block is

μ = \frac{1}{\sqrt{3}}

. The frictional force acting on the block is

Q. As given in the figure two blocks

A

B

of weight

20 N

and

100 N

respectively are in equilibrium. These are being pressed against a wall by force

F

as shown. If the coefficient of friction between the two blocks is

0.1

and between block

B

and the wall is

0.15

, the frictional force applied by wall on block

B

(f_{2})

and the frictional force applied by block

B

on block

A

(f_{1})

Q. Given in the figure are two blocks A and B of weight

20 N

and

100 N

respectively. The blocks are being pressed against a wall by a force

F

as shown. If the coefficient of friction between the blocks is

0.1

and between block B and the wall is

0.15

, the frictional force applied by the wall on block B is

Q. In the figure given below, there are two blocks A and B of weight

20 N

and

100 N

respectively. These are being pressed against a wall by an external force

F

as shown in figure. If the coefficient of static friction between the blocks is

0.1

and between block B and the wall is

0.15

, then the frictional force applied by the wall on block B will be

Q. A block of mass

5 k g

is placed on horizontal surface and a pushing force

20 N

is acting on back as shown in figure. If coefficient of friction between block and surface is

0.2

, calculate frictional force an speed of block after

15 s

Join BYJU'S Learning Program

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.