If accelaration of A is $2 m / s^{2}$ which is smaller than accelaration of B then the value of frictional force applied byon A is

Open in App

Solution

Solution
REF.Image
The ABD of the system will be as: $\to a_{A} = 2 m / s^{2}$
Then,
$f_{s} = m_{A} a_{A}$
$f - f_{s} = m_{B} a_{B}$
As $m_{A} = 5 k g$ and $a_{A} = 2 m / s^{2}$
$f_{s} = (5) (2) = 10 N$
So, frictional force applied
will be 10 N on A.
Note : $f_{s}^{'}$ can be equal or greater
then $f_{s}$ depending upon the surface.

Suggest Corrections

Similar questions

Q. If acceleration of

A

2 m / s^{2}

which is smaller than acceleration of

B

then the value of frictional force applied by

B

A

is:

Q. Block A has a mass of

30 kg

and block B a mass of

15 kg

. The coefficients of friction between all surfaces of contact are

μ_{s} = 0.15

and

μ_{k} = 0.10

. Knowing that

θ = 30^{\circ}

and that the magnitude of the force

F

applied to block A is

250 N

, determine the accelaration of block A.

[Take

g = 10 {m/s}^{2}

]

A body with mass $100 k g$ and volume $0.2 m^{3}$ is taken to a depth of 10 meters in water and released. Density of body is less than water. Accelaration of body after it is released is ___.

Accelaration due to gravity $10 m / s^{2}$

Density of water $1000 k g / m^{3}$ )

Join BYJU'S Learning Program

If accelaration of A is 2m/s2 which is smaller than accelaration of B then the value of frictional force applied byon A is

SolutionREF.ImageThe ABD of the system will be as: →aA=2m/s2Then,fs=mAaAf−fs=mBaBAs mA=5kg and aA=2m/s2fs=(5)(2)=10NSo, frictional force appliedwill be 10 N on A.Note : f′s can be equal or greaterthen fs depending upon the surface.

If accelaration of A is $2 m / s^{2}$ which is smaller than accelaration of B then the value of frictional force applied byon A is