Question

A constant retarding force of 20N acts on a body of mass 5 kg moving initially with a
speed of 10 ms How long does the body take to stop?​

Answer 1

★ Given :

Constant retarding force = 20 N

Mass of body = 5 kg

initial velocity (u) = 10 m/s

final velocity (v) = 0

★ To find :

time taken for body to stop.

★ Solution :

We know that,

F = ma

$a = \frac{v - u}{t}$

$F = m \times \frac{v - u}{t}$

by putting the values

$- 20 = 5 \times \frac{(0 - 10)}{t}$

$- 20 = \frac{-50}{t}$

by cross multiplication

-20 t = -50

$\frac{-50}{-20} = 2.5 \: s$

therefore , time taken = 2.5 seconds..

Answer 2

Given :

• Constant retarding force is 20N.
• Mass is 5 kg.
• Initial velocity (aslo indicated as 'u') is 10 m/s.
• Final velocity (also indicated as 'v') is 0.

To find :

• Time taken by the body to stop.

Formula used :

• F = ma
• $\sf {a = \frac{v - u}{t} }$
• $\sf {F = m \times \frac{v - u}{t} }$

According to the question,

$\sf{ - 20 = 5 \times \frac{0 - 10}{t} }$

$\sf{ - 20 = \frac{ - 50}{t} }$

$\sf{ - 20 \: t = - 50}$

$\sf{t = \frac{ - 50}{ - 20} }$

$\sf{t = \frac{ \cancel - 50}{ \cancel - 20 } = 2.5 \: s }$

• So,the time taken by the body to stop is 2.5 seconds....