Answer
Verified
91.5k+ views
Hint So to solve this type of problem it is necessary to draw the free fall diagram and in this question, we will consider both the blocks are independent and then we will find the acceleration for both the blocks by using the formula$a = g\sin \theta - \mu g\cos \theta $.
Formula used
Acceleration for the block will be
$a = g\sin \theta - \mu g\cos \theta $
Here,
$a$, will be the acceleration of the block
$g$, will be the acceleration due to gravity
Complete Step by Step Solution
Let's consider the two blocks $1$ and$2$, both are independent. Since the blocks are independent so their acceleration will also be independent for both of the blocks.
Acceleration for block$1$:
As we know the formula,
${a_1} = g\sin \theta - {\mu _1}g\cos \theta $
Now substitute the values, we get
$ \Rightarrow g\left[ {\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{2} \times \dfrac{1}{2}} \right]$
On solving the above equation, we get
$ \Rightarrow g\left[ {\dfrac{{2\sqrt 3 - 1}}{4}} \right]$
Therefore, $g\left[ {\dfrac{{2\sqrt 3 - 1}}{4}} \right]$ will be the acceleration for the first block.
Similarly,
Acceleration for block$2$:
As we know the formula,
${a_2} = g\sin \theta - {\mu _2}g\cos \theta $
Now substitute the values, we get
$ \Rightarrow g\left[ {\dfrac{{\sqrt 3 }}{2} - \dfrac{2}{5} \times \dfrac{1}{2}} \right]$
On solving the above equation, we get
$ \Rightarrow g\left[ {5\sqrt 3 - 2} \right]$
Therefore, $g\left[ {5\sqrt 3 - 2} \right]$ will be the acceleration for the second block.
So, we see that the acceleration ${a_2}$is greater than the acceleration of ${a_1}$
Therefore, we can say that both the blocks will move separately.
Hence the option $B$will be correct.
Note Acceleration is positive in the “down” direction, where “down” is where the nearest big source of gravity is.
In physics, acceleration is usually expressed as a vector, i.e. a direction and a magnitude. A magnitude is an absolute value, hence always positive.
We could say the object is accelerating negatively in the “up” direction but that’s a muddled way of looking at things. Direction-and-magnitude makes more sense.
Formula used
Acceleration for the block will be
$a = g\sin \theta - \mu g\cos \theta $
Here,
$a$, will be the acceleration of the block
$g$, will be the acceleration due to gravity
Complete Step by Step Solution
Let's consider the two blocks $1$ and$2$, both are independent. Since the blocks are independent so their acceleration will also be independent for both of the blocks.
Acceleration for block$1$:
As we know the formula,
${a_1} = g\sin \theta - {\mu _1}g\cos \theta $
Now substitute the values, we get
$ \Rightarrow g\left[ {\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{2} \times \dfrac{1}{2}} \right]$
On solving the above equation, we get
$ \Rightarrow g\left[ {\dfrac{{2\sqrt 3 - 1}}{4}} \right]$
Therefore, $g\left[ {\dfrac{{2\sqrt 3 - 1}}{4}} \right]$ will be the acceleration for the first block.
Similarly,
Acceleration for block$2$:
As we know the formula,
${a_2} = g\sin \theta - {\mu _2}g\cos \theta $
Now substitute the values, we get
$ \Rightarrow g\left[ {\dfrac{{\sqrt 3 }}{2} - \dfrac{2}{5} \times \dfrac{1}{2}} \right]$
On solving the above equation, we get
$ \Rightarrow g\left[ {5\sqrt 3 - 2} \right]$
Therefore, $g\left[ {5\sqrt 3 - 2} \right]$ will be the acceleration for the second block.
So, we see that the acceleration ${a_2}$is greater than the acceleration of ${a_1}$
Therefore, we can say that both the blocks will move separately.
Hence the option $B$will be correct.
Note Acceleration is positive in the “down” direction, where “down” is where the nearest big source of gravity is.
In physics, acceleration is usually expressed as a vector, i.e. a direction and a magnitude. A magnitude is an absolute value, hence always positive.
We could say the object is accelerating negatively in the “up” direction but that’s a muddled way of looking at things. Direction-and-magnitude makes more sense.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
The vapour pressure of pure A is 10 torr and at the class 12 chemistry JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
3 mole of gas X and 2 moles of gas Y enters from the class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Derive an expression for maximum speed of a car on class 11 physics JEE_Main
Velocity of car at t 0 is u moves with a constant acceleration class 11 physics JEE_Main