Question

a constant retarding force of 50N. is applied to a body of mass 10 kg moving initially with a speed of 10 m/s the body comes to rest after how many second

Answer 1

here f = - 50 N ( - ve sign refers to retardation)
m= 10 kg ,u=10 m/s -1 ,v=0
f= ma
s= f/m = -50N/ 10 kg
=-5m/s - square
using
v= u+at

t= v-u/ a
0-10/-5 = 2s


so we can conclude that after 2 seconds it will come to rest




# be brainly

Answer 2

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Retarding force,

F = –50 N

Mass of the body,

m = 10 kg

Initial velocity of the body,

u = 10m/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 = 10 × a

=> a= -50/10 = -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 = -10/-5= 2sec.

I hope, this will help you

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