Question

1 A body of mass 20kg and exerting a retarding
force of - 50 N and comes to rest after covering
a certain distance with an initial velocity of 20m/s.
Calculate the distance covered by the body.​

Answer 1

Given :

▪ Mass of body = 20kg

▪ Retarding force = -50N

▪ Initial velocity = 20mps

▪ Final velocity = zero (i.e., rest)

To Find :

▪ Distance covered by body in the given interval of motion.

Concept :

❇ Force is defined as the product of mass and acceleration.

❇ Here, Retarding force has said to be constant (constant retardation) throughout the motion, we can easily apply equation of kinematics to solve this question.

❇ As per Newton's second law of motion,

$\bigstar\:\underline{\boxed{\bf{\red{F=m\times a}}}}$

❇ Third equation of kinematics,

$\bigstar\:\underline{\boxed{\bf{\blue{v^2-u^2=2as}}}}$

Calculation :

$\dashrightarrow\sf\:v^2-u^2=2as\\ \\ \dashrightarrow\sf\:v^2-u^2=2s\times \dfrac{F}{m}\\ \\ \dashrightarrow\sf\:(0)^2-(20)^2=2s\times \dfrac{(-50)}{20}\\ \\ \dashrightarrow\sf\:-400=-5s\\ \\ \dashrightarrow\sf\:s=\dfrac{-400}{-5}\\ \\ \dashrightarrow\underline{\boxed{\bf{\orange{s=80m}}}}\:\gray{\bigstar}$

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

• Mass(m)=20kg
• Retardation(f)=-50N
• Intial velocity(u)=20m/s

${\bf{\blue{\underline{To\:Find:}}}}$

• Distance(s)=?

${\bf{\blue{\underline{Now:}}}}$

${\implies{\sf{ \: f = m \times a}}}$

${\implies{\sf{ \: a = \frac{f}{m} }}}$

According to the question,

${\implies{\sf{ \: {v}^{2} - {u}^{2} = 2as}}}$

${\implies{\sf{ \: {(0)}^{2} - {(20)}^{2} = 2s \frac{f}{m} }}}$

${\implies{\sf{ \: {(0)}^{2} - {(20)}^{2} = 2s \frac{ - 50}{20} }}}$

${\implies{\sf{ \: - 400 = s \frac{ - 50}{10} }}}$

${\implies{\sf{ \: - 400 = - 5s }}}$

${\implies{\sf{ \: s = \frac{400}{5} }}}$

${\implies \boxed{\sf{ \: s = 80m }}}$