Question

retarding force of 50n is applied to a body of mass 20g moving with a speed of 15m/s . How long would it take to stop.

Answer 1

hey
here's your answer!!

F = - 50 N
m = 20 g = 20/1000 kg
u = 0
v = 15 m/s
t = ?

now, we know that
F = m.a
=> F = m. (v-u)/t
=> - 50 N = 20 kg/1000 * (0-15)m/s /t
=> - 50 N * t = 20 kg/1000 * -15 m/s
=> 50 N * t = 20 kg/1000 * 15 m/s
=> t * 50 N = 300/1000 kg m/s
=> t = 6/1000 s
=> t = 0.006 s (answer)

Answer 2

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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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