Question

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 .​

Answer 1

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

$\rule{300}{1.5}$