Question

A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s⁻¹. How long does the body take to stop?

Answer 1

As we know from Newton's second law,


F = ma

Given F is retarding force,

This means :-

=> Force is negative

=> acceleration is negative

So given

F = - 50N
mass = 20kg
a = ??

From,

F = ma

=> (-50) = 20a

=> a = -50/20

=> a = -5/2 m/s²

But we know that acceleration is the rate of change of velocity

=> a = (v - u)/t


When the body will stop, v will become zero.
and we are given that u is 15 m/s
acceleration as we derived above is -5/2 m/s²

=> a = (v - u)/t

=>
$\frac{ - 5}{2} = \frac{0 - 15}{t} \\ \\ = > \frac{ - 5}{2} = \frac{ - 15}{t} \\ \\ = > - 5t = - 15 \times 2 \\ \\ = > t = \frac{ - 15 \times 2}{ - 5} \\ \\ = > t = 6$
Hence after 6 seconds, the final velocity will be zero, that is, after 6 seconds, the body will stop.


Hope it helps dear friend ☺️✌️

Answer 2

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$\bold\red{hello...frd\:swigy\:here}$

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Retarding force,

F = –50 N

Mass of the body,

m = 20 kg

Initial velocity of the body,

u = 15 m/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

=> –50 = 20 × a

=> a= -50/20 = -2.5 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 = -15/-2.5= 6 sec.

I hope, this will help you

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