Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Answer please! A block of mass m is connected to another block of mass M by a spring ( massless ) of spring constant K . The blocks are kept on a smooth horizontal plane.Initially the blocks are at rest and the spring is unstretched Then a constan

A block of mass m is connected to another block of mass  M  by a spring ( massless ) of spring constant  K . The blocks are kept on a smooth horizontal plane.Initially the blocks are at rest and the spring is unstretched   Then a constant force  F  starts acting on the block of mass  M  to pull it. Find the force of the block of mass  m.

 

  • Option 1)

    \frac{MF}{\left ( m+M \right )}

  • Option 2)

    \frac{mF}{M}

  • Option 3)

    \frac{\left ( M+m \right )F}{m}

  • Option 4)

    \frac{mF}{(m+M)}

 

Answers (1)

best_answer

As we learnt in

When 2 Blocks are in Contact -



F-f=m_{1}a

f=m_{2}a

- wherein

a=\frac{F}{m_{1}+m_{2}}

f=\frac{m_{2}F}{m_{1}+m_{2}}

 

 

 

: Acceleration \: of\: the\: system\: a= \frac{F}{m+M}

Force \: on \: block\: of \: mass\: m= ma= \frac{mF}{m+M}

Correct option is 4.


Option 1)

\frac{MF}{\left ( m+M \right )}

This is an incorrect option.

Option 2)

\frac{mF}{M}

This is an incorrect option.

Option 3)

\frac{\left ( M+m \right )F}{m}

This is an incorrect option.

Option 4)

\frac{mF}{(m+M)}

This is the correct option.

Posted by

prateek

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Advanced 2024: Chemical engineering cut-offs at IITs; know opening, closing ranks
anam-khan

IIT JEE Mains Result 2024 Session 2: How is overall merit list prepared?
anam-khan

JEE Mains Session 2 result 2024 soon; List of top NITs in India

Latest Question