You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Physics

Second Law of Motion

Consider 2 bl...

Question

Consider 2 blocks with masses $m_{1} = m_{2}$ = 6 kg. A force F= 20 N acts on them and causes an acceleration of $0.5 m / s^{2}$ . If the friction forces on the two blocks are equal and in same direction. What is the magnitude of frictional force on either block?

14.0 N

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

17.0N

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

7.0 N

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

Open in App

Solution

The correct option is C 7.0 N
$a = \frac{F - 2 f_{k}}{m}$
$\Rightarrow F - 2 f_{k} = 12 \times 0.5 \Rightarrow f_{k} = 7.0 N$

Suggest Corrections

Similar questions

Q. In the shown figure. If

F = 20 N, m_{1} = m_{2} = 3 k g

and the acceleration is

0.5 m / s^{2}

. What will be the tension in the connecting cord if the friction forces on the two blocks are equal? How large is the frictionless force on either block.

Consider two blocks of masses m₁and m₂ placed in contact with each other on a friction less horizontal surface. Let a force F be applied on block of mass m₁, then the value of contact force between the blocks is

Q. Two blocks of masses

m_{1} = 5 k g

and

m_{2} = 6 k g

are connected by a light string passing over a light frictionless pulley as shown. The mass

m_{1}

is at rest on the inclined plane and mass

m_{2}

hangs vertically. If the angle of incline

θ = 30^{0}

what is magnitude and direction of force of friction on the 5kg block

(g = 10 m s^{- 2})

Two blocks of equal masses $m_{1} = m_{2} = 3 k g$ , connected by a light string, are placed on a horizontal surface which is not frictionless. If a force of 20 N is applied in the horizontal direction on a block, the acceleration of each block is $0.5 m s^{- 2}$ . Assuming that the frictional forces on the two blocks are equal, the tension in the string will be

Q. Two trolleys 1 and 2 are moving with accelerations

a_{1}

and

a_{2}

, respectively, in the same direction. A block of mass m on trolley 1 is in equilibrium from the frame of observer stationary with respect to trolley

2

. The magnitude of friction force on block due to trolley is (assume that no horizontal force other than friction force is acting on block).

Join BYJU'S Learning Program

Consider 2 blocks with masses m1=m2= 6 kg. A force F= 20 N acts on them and causes an acceleration of 0.5 m/s2. If the friction forces on the two blocks are equal and in same direction. What is the magnitude of frictional force on either block?

The correct option is C 7.0 Na=F−2fkm ⇒F−2fk=12×0.5⇒fk=7.0 N

Consider 2 blocks with masses $m_{1} = m_{2}$ = 6 kg. A force F= 20 N acts on them and causes an acceleration of $0.5 m / s^{2}$ . If the friction forces on the two blocks are equal and in same direction. What is the magnitude of frictional force on either block?

The correct option is C 7.0 N
$a = \frac{F - 2 f_{k}}{m}$
$\Rightarrow F - 2 f_{k} = 12 \times 0.5 \Rightarrow f_{k} = 7.0 N$