Question

A body with mass 30 kg with speed 20m/s applies force 60n and how much does it takes to stop

Answer 1

Stopping distance =
$mu {}^{2} \div 2force$
30×20×20/2×60
=30×400/120
=1200/12
=100m

Stopping time = mu/force
=30×20/60
=600/60
=100 sec

Answer 2

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

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Retarding force,

F = –60 N

Mass of the body,

m = 30 kg

Initial velocity of the body,

u = 20 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

=> –60 = 30 × a

=> a= -60/30 = -2 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/-2= 15 sec.

I hope, this will help you

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