CameraIcon

SearchIcon

MyQuestionIcon

1

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

Spring Force

Two blocks of...

Question

Two blocks of masses $m_{1}$ and $m_{2}$ are connected with a light spring of force constant $k$ and the whole system is kept on a frictionless horizontal surface. The masses are applied forces $F_{1}$ and $F_{2}$ as shown in figure. At any time, the blocks have same acceleration $a_{0}$ but in opposite directions.
Then acceleration $a_{0}$ is

A

\frac{F_{1} - F_{2}}{m_{1} + m_{2}}

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

B

\frac{F_{1} - F_{2}}{m_{1} - m_{2}}

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

C

\frac{F_{1} + F_{2}}{m_{1} - m_{2}}

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

D

\frac{F_{1} + F_{2}}{m_{1} + m_{2}}

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

Open in App

Solution

The correct option is B $\frac{F_{1} - F_{2}}{m_{1} - m_{2}}$
FBD

Using Newtons second law $F_{n e t} = m_{1} a_{0} = F_{1} - k x$

Using Newtons second law $F_{n e t} = F_{2} - k x = m_{2} a_{0}$ (as per the question $a_{2} = a_{0}$ )

Subtracting the equations: $m_{1} a_{0} - m_{2} a_{0} = F_{1} - F_{2}$
$\Rightarrow a_{0} = \frac{F_{1} - F_{2}}{m_{1} - m_{2}}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Two blocks of masses

m_{1}

m_{2}

are connected with a light spring of force constant

k

and the whole system is kept on a frictionless horizontal surface. The masses are applied forces

F_{1}

F_{2}

as shown in figure. At any time, the blocks have same acceleration

a_{0}

but in opposite directions.
Then acceleration

a_{0}

F_{1}

m_{1}

towards east and

F_{2}

m_{2}

towards west, then the contact force between

m_{1}

m_{2}

(F_{1} > F_{2})

Q. Three blocks

m_{1}, m_{2}

m_{3}

8 kg, 3 kg

1 kg

are placed in contact on a smooth surface. Forces

F_{1} = 140 N

F_{2} = 20 N

are acting on blocks

m_{1}

m_{3}

respectively as shown. The reaction between blocks

m_{2}

m_{3}

Q. Three blocks

m_{1}, m_{2}

m_{3}

8 kg, 3 kg

1 kg

are placed in contact on a smooth surface. Forces

F_{1} = 140 N

F_{2} = 20 N

are acting on blocks

m_{1}

m_{3}

respectively as shown. The reaction between blocks

m_{2}

m_{3}

Q. Four blocks of the same material,

m_{1}, m_{2}, m_{3}

m_{4}

2 k g, 4 k g, 2.5 k g

2 k g

respectively. All of them have equal surface area in contact with the concrete surface over which they are kept. Let the frictional force acting on them be

F_{1}, F_{2}, F_{3}

F_{4}

respectively.
The correct order of friction force acting on them when they slide is:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Springs: Playing with a Slinky

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon