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

Byju's Answer

Standard XII

Physics

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

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

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

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

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

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

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

\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

Similar questions

Q. 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}

Q. 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}

Q. Two blocks of masses

m_{1} = 1 k g

and

m_{2} = 2 k g

are connected by massless strings and placed on a horizontal frictionless surface as shown in figure. A force

F = 12 N

is applied to mass

m_{1}

as shown then tension

(T)

between

m_{1}

and

m_{2}

Q. Three blocks of masses

M_{1}, M_{2}

and

M

are arranged as shown in the figure. All the surfaces are frictionless and string is inextensible. Pulleys are light. A constant force

F

is applied on block of mass

M_{1}

. Pulleys and string are light. Part of the string connecting both pulleys is vertical and part of the string connecting pulleys with masses

M_{1} and M_{2}

are horizontal.

(P) Acceleration of mass $M_{1}$	(1) $\frac{F}{M_{2}}$
(Q) Acceleration of mass $M_{2}$	(2) $\frac{F}{M_{1} + M_{2}}$
(R) Acceleration of mass $M$	(3) zero
(S) Tension in the string	(4) $\frac{M_{2} F}{M_{1} + M_{2}}$

Three blocks of masses $m_{1} = 1 k g, m_{2} = 2 k g a n d m_{3} = 3 k g$ are placed in contact on a horizontal frictionless surface. A force $F = 12 N$ is applied to mass $m_{1}$ . The acceleration of the system is

Join BYJU'S Learning Program

The correct option is B F1−F2m1−m2FBD Using Newtons second law Fnet=m1a0=F1−kx Using Newtons second law Fnet=F2−kx=m2a0 (as per the question a2=a0) Subtracting the equations: m1a0−m2a0=F1−F2 ⇒a0=F1−F2m1−m2