Question

Force acting on a body of mass is 700 gm and is 0.49N is brought to rest in 6 sec then what will be distance covered from body

Answer 1

Answer:

• Distance Covered (S) is 12.6 Meters.

Given:

1. Force Acting (F) = - 0.49 N.
2. Mass of the Body (M) = 700 grams = 0.7 Kg.
3. Time Period (t) = 6 Seconds.
4. Final velocity (v) = 0 m/s

Explanation:

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From Newtons Second Law.

$\large\star \; {\boxed{\bold{F = Ma}}}$

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

(Here we need to take Force value as Negative as the Force is retarding.)

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

Substituting the Values.

$\large{\tt \hookrightarrow - 0.49 \; N = 0.7 \; Kg \times a}$

$\large{\tt \hookrightarrow - 0.49 = 0.7 \times a}$

$\large{\tt \hookrightarrow a = \dfrac{ - 0.49}{0.7}}$

$\large{\tt \hookrightarrow a = \dfrac{ - 49 \times 10}{7 \times 100}}$

$\large{\tt \hookrightarrow a = \cancel{\dfrac{ - 49 \times 10}{7 \times 100}}}$

$\large{\tt \hookrightarrow a = \dfrac{ - 7}{10}}$

$\large{\hookrightarrow {\underline{\underline{\tt a = 0. 7 \; m/s^2}}}}$

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From First Kinematic equation

$\large\star \: {\boxed{\bold{v = u +at}}}$

$\bold{Here}\begin{cases}\text{v Denotes Final velocity} \\ \text{u denotes Initial velocity} \\ \text{a Denotes Acceleration} \\ \text{t Denotes Time taken}\end{cases}$

$\large{\boxed{\tt v = u + at}}$

Substituting the values.

$\large{\tt \hookrightarrow 0 = u + (- 0.7) \times 6}$

$\large{\tt \hookrightarrow 0 = u - 0.7 \times 6}$

$\large{\tt \hookrightarrow u = 0.7 \times 6}$

$\large{\hookrightarrow {\underline{\underline{\tt u = 4.2 \; m/s}}}}$

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From Third Kinematic Equation.

$\large\star \: {\boxed{\bold{v^2 = u^2 + 2as}}}$

$\bold{Here}\begin{cases}\text{v Denotes Final velocity} \\ \text{u denotes Initial velocity} \\ \text{a Denotes Acceleration} \\ \text{s Denotes Distance Travelled}\end{cases}$

$\large{\boxed{\tt v^2 = u^2 + 2as}}$

Substituting the Values,

$\large{\tt \hookrightarrow (0)^2 = (4.2)^2 + 2 \times ( - 0.7) \times s}$

$\large{\tt \hookrightarrow 0 = 17.64 + 2 \times ( - 0.7) \times s}$

$\large{\tt \hookrightarrow 0 - 17.64 = 2 \times ( - 0.7) \times s}$

$\large{\tt \hookrightarrow - 17.64 = - 1.4 \times s}$

$\large{\tt \hookrightarrow \cancel{ -} 17.64 = \cancel{-} 1.4 \times s}$

$\large{\tt \hookrightarrow 17.64 = 1.4 \times s}$

$\large{\tt \hookrightarrow s = \dfrac{17.64}{1.4}}$

$\large{\tt \hookrightarrow s =\cancel{ \dfrac{17.64}{1.4}}}$

$\huge{\boxed{\boxed{\tt s = 12.6 \; m}}}$

∴ Distance Covered By the Body is 12.6 Meters.

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