Question

can anyone solve this ASAP​

Answer 1

Answer:

Because the acceleration of the rope is of the same magnitude at every point in the rope, the acceleration of the two masses will also be of equal magnitude. If we label the acceleration of mass m as a , then the acceleration of mass M is a. Using Newtons Second Law we find: 

For mass M : Mg−T=ma

for mass m : T−mg=ma

By subtracting the first equation from the second, we find (M−m)g=(M+m)a

$= > a = \frac{m - m}{m + m} g$

Because M−m>0 , a is positive and mass m accelerates upward as anticipated. This result gives us a general formula for the acceleration of any pulley system with unequal masses, M and m . Remember, the acceleration is positive for m and negative for M , since m is moving up and M is going down.

Answer 2

Question :

The given figure the acceleration of mass M is $a_{M} = 2ms {}^{ - 2}$

then acceleration of the mass m would be :

Formula Uesd :

In the pulley mass systems problems:

${\purple{\boxed{\large{\bold{at=constant}}}}}$

Where a = acceleration

T= tension

Solution :

$a_{m} = 2\sqrt{10}\:ms{}^{ - 2}$

correct option 4)

Explanation :

Given :$a_{M} = 2ms {}^{ - 2}$

Let the tension created by mg force be t

since the string is continuous , so the tension in all over the string is t.

Now draw FBD to understand clearly the Question:

we observed that on Mass m

• 3t force act in horizontal direction
• And t in vertical ( downward direction)

Resultant of forces, $t_{M}$=t√10

Now apply pully mass system condition:

$a_{M} \times t _{M} = a_{m} \times t_{m}$

$\implies 2 \times t \times \sqrt{10} = a_{m} \times t$

$\implies \: a _{m} = 2 \times </p><p> \sqrt{10} \: ms {}^{ - 2}$