Question

two bodies have a mass of ratio 4:5 the force applied on the bigger mass produce an a acceleration of 15 m/s². if the same force act on it, calculate the acceleration on the lighter mass?,​

Answer 1

Given:-

• Ratio of mass of two bodies M1 : M2 = 4:5
• Constant force is applied on both bodies
• Acceleration produce in 1st body A1 = 15m/s²

To Find:-

• Acceleration produce in 2nd body ?

Solution:-

⠀⠀⠀⠀⠀⠀⠀⠀According to the Given Statement

We have to calculate the acceleration produce in 2nd body . As we know that Force is defined as the product of mass & acceleration.

Let the acceleration of 2nd body = A2

It is also given that constant force is applied on both bodies . It means the

• F1 = F2 (both are equal )

where,

• F1 denote Force on 1st body
• F2 denote force on 2nd body

Now,

Substitute the value we get

$:\implies$ M1A1 = M2A2 (product of Mass & acceleration)

$:\implies$ 4×15 = 5× A2

$:\implies$ 60 = 5 × A2

$:\implies$ 60/5 = A2

$:\implies$ A2 = 60/5

$:\implies$ A2 = 12m/s²

• Hence, the acceleration on the lighter mass is 12m/s².

Answer 2

Answer:

Given :-

• Two bodies have a mass of ratio of 4 : 5.
• The force applied on the bigger mass produce an acceleration of 15 m/s².
• The same force act on it.

To Find :-

• What is the acceleration of the lighter mass.

Formula Used :-

$\clubsuit$ Force Formula :

$\longmapsto \sf\boxed{\bold{\pink{F = ma}}}$

where,

• F = Force
• m = Mass
• a = Acceleration

Solution :-

Let,

$\mapsto$ The acceleration of the lighter mass = a₂

Given :

• First mass (m₁) = 4
• Second mass (m₂) = 5
• First acceleration (a₁) = 15 m/s²

According to the question by using the formula we get,

$\dashrightarrow \sf\bold{F_1 =\: F_2}$

$\implies \sf m_1a_1 =\: m_2a_2$

$\implies \sf 4 \times 15 =\: 5 \times a_2$

$\implies \sf 60 =\: 5a_2$

$\implies \sf a_2 =\: \dfrac{\cancel{60}}{\cancel{5}}$

$\implies \sf a_2 =\:\dfrac{12}{1}$

$\implies \sf\bold{\red{a_2 =\: 12\: m/s^2}}$

$\therefore$ The acceleration of the lighter mass is 12 m/s².