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

Limiting Friction

In the follow...

Question

In the following figure, two blocks of $m_{1} = 20 k g$ and $m_{2} = 10 k g$ are in contact with a frictionless table. A horizontal force F=3N is applied to mass $m_{1}$ the contact force between $m_{1}$ and $m_{2}$ will be

1 N

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

2 N

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

3 N

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

zero

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

Open in App

Solution

The correct option is A $1 N$
Let the contact force be $N$
$F = (m_{1} + m_{2}) a = 30 a = 3$
$a = 0.1 m / s^{2}$
Now,
The contact force is the reason for the acceleration of $m_{2}$
$N = m_{2} \times a$
$N = 10 \times 0.1 = 1$
Option $A$ is the correct answer

Suggest Corrections

Similar questions

Q. Three blocks of masses

m_{1}, m_{2}

and

m_{3}

are lying in contact with each other on a horizontal frictionless plane as shown in the figure. If a horizontal force

F

is applied on

m_{1}

then the force due to

m_{2}

m_{1}

will be:

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. A single horizontal force F is applied to a block of mass

M_{1}

. Mass

M_{1}

is in contact with another block of mass

M_{2}

. If the surfaces are frictionless, the force between the blocks is:

Q. A force

F

is applied horizontally on mass

m_{1}

as shown in figure. Find the contact force between

m_{1}

and

m_{2}

Q. A single horizontal force F is applied to a block of mass

M_{1}

,which is in contact with another block of mass

M_{2}

(Fig. 3.2). If the surfaces are frictionless, the force between the blocks is

Join BYJU'S Learning Program

In the following figure, two blocks of m1=20kg and m2=10kg are in contact with a frictionless table. A horizontal force F=3N is applied to mass m1 the contact force between m1 and m2 will be

The correct option is A 1NLet the contact force be NF=(m1+m2)a=30a=3a=0.1m/s2Now,The contact force is the reason for the acceleration of m2N=m2×aN=10×0.1=1Option A is the correct answer

In the following figure, two blocks of $m_{1} = 20 k g$ and $m_{2} = 10 k g$ are in contact with a frictionless table. A horizontal force F=3N is applied to mass $m_{1}$ the contact force between $m_{1}$ and $m_{2}$ will be

The correct option is A $1 N$
Let the contact force be $N$
$F = (m_{1} + m_{2}) a = 30 a = 3$
$a = 0.1 m / s^{2}$
Now,
The contact force is the reason for the acceleration of $m_{2}$
$N = m_{2} \times a$
$N = 10 \times 0.1 = 1$
Option $A$ is the correct answer