Question

A force produces an acceleration of 16 m/ s in a body of mass 0.5 kg and an acceleration of 4.0 m/ s in another body .if both the bodies are fastened together then how much acceleration will be produced by this force.

Answer 1

Correct Question:

A force produces an acceleration of 16 m/s² in a body of mass 0.5 kg and an acceleration of 4.0 m/s² in another body .If both the bodies are fastened together then how much acceleration will be produced by this force.

Answer:

• The Acceleration (a) will be 3.2 m/s².

Given:

1. Mass of First Body (M₁) = 0.5 Kg.
2. Acceleration of First body (a₁) = 16 m/s².
3. Acceleration of Second body (a₂) = 4 m/s².

Explanation:

$\rule{300}{1.5}$

Applying Newton's Second law of motion,

★ $\large{\boxed{\sf{F = Ma}}}$

$\bold{Here}\begin{cases}\text{F Denotes Force} \\ \text{M Denotes Mass} \\ \text{a Denotes Acceleration}\end{cases}$

Now, For First Body,

$\large{\boxed{\tt F = M_1a_1}}$

Substituting the values,

$\large{\tt \hookrightarrow F = 0.5 \times 16 }$

$\large{\tt \hookrightarrow F = \dfrac{5}{10} \times 16 }$

$\large{\tt \hookrightarrow F = \cancel{\dfrac{5}{10}} \times 16 }$

$\large{\tt \hookrightarrow F = \dfrac{1}{2} \times 16 }$

$\large{\tt \hookrightarrow F = \cancel{\dfrac{16}{2}}}$

$\large{\underline{\boxed{\tt F = 8 \: N}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Applying Newton's Second law of motion,

★ $\large{\boxed{\sf{F = Ma}}}$

$\bold{Here}\begin{cases}\text{F Denotes Force} \\ \text{M Denotes Mass} \\ \text{a Denotes Acceleration}\end{cases}$

Now, For Second Body

$\large{\boxed{\tt F = M_2a_2}}$

Force Remains same.

$\large{\tt \hookrightarrow 8 \: N = M_2 \times 4 \: m/s^2}$

$\large{\tt \hookrightarrow 8 = M_2 \times 4}$

$\large{\tt \hookrightarrow M_2 = \dfrac{8}{4}}$

$\large{\tt \hookrightarrow M_2 = \cancel{\dfrac{8}{4}}}$

$\large{\underline{\boxed{\tt M_2 = 2 \: Kg}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Applying Newton's Second law of motion,

★ $\large{\boxed{\sf{F = Ma}}}$

$\bold{Here}\begin{cases}\text{F Denotes Force} \\ \text{M Denotes Mass} \\ \text{a Denotes Acceleration}\end{cases}$

Now, For Combined Mass,

$\large{\boxed{\tt F = M'a}}$

Force will remain Same.

Substituting the values,

$\large{\tt \hookrightarrow F = (M_1 + M_2) \times a}$

$\large{\tt \hookrightarrow 8 \: N = (0.5 + 2) \; Kg \times a}$

$\large{\tt \hookrightarrow 8 = (0.5 + 2) \times a}$

$\large{\tt \hookrightarrow 8 = 2.5 \times a}$

$\large{\tt \hookrightarrow a = \dfrac{8}{2.5}}$

$\large{\tt \hookrightarrow a = \dfrac{80}{25}}$

$\large{\tt \hookrightarrow a = \cancel{\dfrac{80}{25}}}$

$\huge{\boxed{\boxed{\tt a = 3.2 \: m/s^2}}}$

∴ Acceleration (a) of Combined Mass is 3.2 m/s².

$\rule{300}{1.5}$