Question

Two blocks of mass m1 = 1 kg and m2 = 2 kg are placed in contact on a friction less horizontal surface. A force of 10 N is acting on m1. What is the acceleration of m1 and m2. What is the magnitude of action and reaction?

Answer 1

Given :

▪ Mass of block A = 1kg

▪ Mass of block B = 2kg

▪ Applied force on block A = 10N

To Find :

▪ Acceleration of the system.

▪ Magnitude of action/reaction force b/w both blocks.

Concept :

→ As per Newton's second law of motion, Ner force acting on the system is defined as the product of mass and acceleration.

→ It is a vector quantity.

→ As per Newton's third law of motion,

Reaction = -(Action )

[Note : Negative sign shows opposite direction.]

Diagram :

☞ Please, see the attachment for better understanding.

Calculation :

✴ Acceleration of the system :

$\red{\twoheadrightarrow}\tt\:F_{net}=(m_1+m_2)\times a\\ \\ \red{\twoheadrightarrow}\tt\:10=(1+2)\times a\\ \\ \red{\twoheadrightarrow}\tt\:a=\dfrac{10}{3}\\ \\ \red{\twoheadrightarrow}\underline{\boxed{\bf{\large{\purple{a=3.33\:ms^{-2}}}}}}\:\orange{\bigstar}$

✴ Action/Reaction force b/w A and B :

$\green{\dashrightarrow}\tt\:F_{net}-R=m_1\times a\\ \\ \green{\dashrightarrow}\tt\:10-R=1\times\dfrac{10}{3}\\ \\ \green{\dashrightarrow}\tt\:R=10-\dfrac{10}{3}\\ \\ \green{\dashrightarrow}\tt\:R=\dfrac{30-10}{3}=\dfrac{20}{3}\\ \\ \green{\dashrightarrow}\underline{\boxed{\bf{\large{\pink{R=6.67\:N}}}}}\:\gray{\bigstar}$

Answer 2

Given :-

m1 = 1 kg

m2 = 2kg

Force on first block = 10 N

To find :-

• The acceleration of m1 and m2 ?

• The magnitude of action and reaction?

Answer :-

➡ Acceleration of m1 & m2 is 3.33 ms-²

➡The magnitude of action and reaction = 6.66N

Solution :-

⭐ Acceleration of m1 & m2,

➡Net Force = (m1+m2) × a

➡10 = ( 1 + 2) × a

➡$a = \dfrac{10}{3}$

✒a = 3.33 ms-² ✅

⭐The magnitude of action and reaction,

➡Net Force - R = m1 × a

➡$10 - R = 1 \times \frac{10}{3}$

➡R = $= 10 - \dfrac{10}{3}$

➡R =$= \dfrac{30 - 10}{3}$

➡R =$= \dfrac{20}{3}$

✒R = 6.67 N ✅

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