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

Hey mate,

# Answer- 6 s

## Explaination-
# Given-
F = -50 N
m = 20 kg
u = 15 m/s
v = 0

# Solution-
Force acting on body is given by,
F = m×a
F = m(v-u)/t
-50 = 20(0-15)/t
t = 20×15/50
t = 6 s

Body will take 6 s to stop due to retarding force.

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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