Physics, asked by fk52716faijankhan, 11 months ago

A constant retarding force of 50 N is applied to
body of mass 10kg moving initially with a speed of
10 ms. when body comes to rest .​

Answers

Answered by ShivamKashyap08
4

Answer:

  • Body will Come to Rest after Time (t) 2 Seconds.

Given:

  1. Retarding Force (F) = 50 N.
  2. Mass of The Body (M) = 10 Kg.
  3. Initial Velocity (u) = 10 m/s.

Explanation:

\rule{300}{1.5}

From Newton's Law of Motion.

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

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

\large{\boxed{\tt F = ma}}

Substituting the Values.

\large{\tt \hookrightarrow - 50 \; N = 10 \; Kg \times a}

Here Force is Negative As the Force is retarding.

\large{\tt \hookrightarrow - 50 = 10 \times a}

\large{\tt \hookrightarrow a = \dfrac{- 50}{10}}

\large{\tt \hookrightarrow a =\cancel{\dfrac{- 50}{10}}}

\large{\underline{\boxed{\tt a = - 5 \; m/s^2}}}

\rule{300}{1.5}

\rule{300}{1.5}

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 = 10 + ( - 5) \times t}

\large{\tt \hookrightarrow 0 = 10 - 5t}

\large{\tt \hookrightarrow 5t = 10}

\large{\tt \hookrightarrow t = \dfrac{10}{5}}

\large{\tt \hookrightarrow t = \cancel{\dfrac{10}{5}}}

\huge{\boxed{\boxed{\tt t = 2 \; Sec}}}

Body will Come to rest After Time (t) 2 seconds

\rule{300}{1.5}

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