Question

A constant retarding force of 100n is applied to a body of mass 10kg moving initially with a speed of 30 ms-1. What is the retardation of the body

Answer 1

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Retarding force,

F = –100 N

Mass of the body,

m = 10 kg

Initial velocity of the body,

u = 30m/s

Final velocity of the body,

v = 0

Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as:

F = ma

=> –100 = 10 × a

=> a= -100/10 = -10 m/s^2

Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as:

v = u + at

=> t= -u/a = -30/-10= 3sec.

I hope, this will help you

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Answer 2

Answer:-

it takes 3 second for body to stop.

Explanation:-

Given:-

Final velocity (v) =0

Initial velocity (u)= 30 m/s

Force( F)= 100 N

Mass of body (m)= 10kg

Solution:-

$\bf{retardation \:(a) = \frac{F}{m} } \\ \\ \implies \: \frac{100}{10} = 10 \: \: \: \frac{m}{ {s}^{2} } \\ \\ \red{ \bf{now \: using \: first \: equation \: of \: motion}} \\ \\ \implies \bf{ v = u + at} \\ \\ \implies \: \bf{ 0 = 30 + ( - 10)t }\\ \\ \implies \: \bf{10t = 30 }\\ \\ \implies \: \bf{t = 3 \: s}$

Hence, it takes 3 second for body to stop.