
Two blocks of masses 7 kg and 3 kg are connected by a spring of stiffness 1000 N/m and placed on a smooth horizontal surface. They are acted by horizontal forces of 72 N and 32N in opposite directions as shown in figure. When accelerations of the blocks are equal and constant, the extensions of the spring is
(A) 4cm
(B) 5cm
(C) 6cm
(D) 8cm
Answer
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Hint: This could be simply solved by breaking the diagrams into simple free body diagrams of both the blocks. Then we need to apply Newton’s law of motion.
Formula used: Here, we will use the basic formula NLM-2:
${\text{F = ma}}$
Here, ${\text{F}}$ is the force exerted on the block
${\text{m}}$ is the mass of the block
${\text{a}}$ is acceleration of the block
Also, spring force = ${\text{kx}}$
Here, ${\text{k}}$ is the spring constant
${\text{x}}$ is the extension due to force
Complete step by step answer:
We already know that the bodies are connected together by rope,
So, both the bodies will have the same acceleration.
First, we will consider the block of mass 3kg:
Let us assume the acceleration of the block to be${\text{a}}$,
There will also be a tension ${\text{T}}$ in the block on cutting the rope with which it is attached.
UsingNLM-2:
${\text{F = ma}}$
${\text{72 - T = 3a}}$
Also, for the block of mass 7kg,
${\text{T - 32 = 7a}}$
On solving, the above equations,
${\text{a = 4m/}}{{\text{s}}^{\text{2}}}$
${\text{T = 60N}}$
Now, we will calculate the extension in the spring:
${\text{T = kx}}$
On putting the values,
${\text{x = }}\dfrac{{\text{T}}}{{\text{k}}}{\text{ = }}\dfrac{{{\text{60}}}}{{{\text{1000}}}}$
$ \Rightarrow {\text{x = }}6 \times {10^{ - 2}}$
So, the extension in the spring is 0.06m or 6cm.
The correct option is C.
Note:
It should be always kept in mind that we should always draw the FBD of the body before solving any force questions.
Also, never forget to assume tension force away from the body wherever there is rope in pulley or block questions.
Similarly, spring forces also are there is case of springs acting tensile (away) from the body.
Also, if the bodies are connected together, they will have the same acceleration.
Formula used: Here, we will use the basic formula NLM-2:
${\text{F = ma}}$
Here, ${\text{F}}$ is the force exerted on the block
${\text{m}}$ is the mass of the block
${\text{a}}$ is acceleration of the block
Also, spring force = ${\text{kx}}$
Here, ${\text{k}}$ is the spring constant
${\text{x}}$ is the extension due to force
Complete step by step answer:
We already know that the bodies are connected together by rope,
So, both the bodies will have the same acceleration.
First, we will consider the block of mass 3kg:
Let us assume the acceleration of the block to be${\text{a}}$,
There will also be a tension ${\text{T}}$ in the block on cutting the rope with which it is attached.
UsingNLM-2:
${\text{F = ma}}$
${\text{72 - T = 3a}}$
Also, for the block of mass 7kg,
${\text{T - 32 = 7a}}$
On solving, the above equations,
${\text{a = 4m/}}{{\text{s}}^{\text{2}}}$
${\text{T = 60N}}$
Now, we will calculate the extension in the spring:
${\text{T = kx}}$
On putting the values,
${\text{x = }}\dfrac{{\text{T}}}{{\text{k}}}{\text{ = }}\dfrac{{{\text{60}}}}{{{\text{1000}}}}$
$ \Rightarrow {\text{x = }}6 \times {10^{ - 2}}$
So, the extension in the spring is 0.06m or 6cm.
The correct option is C.
Note:
It should be always kept in mind that we should always draw the FBD of the body before solving any force questions.
Also, never forget to assume tension force away from the body wherever there is rope in pulley or block questions.
Similarly, spring forces also are there is case of springs acting tensile (away) from the body.
Also, if the bodies are connected together, they will have the same acceleration.
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