A constant force of magnitude 50 n is applied on a body of mass 10 kg moving initially with a speed of 10 ms-1 against the direction of motion. how long will it take the body to come to rest?
Answers
Answered by
1
F= ma
50=10xa
a=5m/s2
the deaccleration is 5.0m/s2
find out time to stop
v=u+at
0=10-5t
t=10/5
t=2sec
50=10xa
a=5m/s2
the deaccleration is 5.0m/s2
find out time to stop
v=u+at
0=10-5t
t=10/5
t=2sec
Answered by
3
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Retarding force,
F = –50 N
Mass of the body,
m = 10 kg
Initial velocity of the body,
u = 10 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 = 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= 2 sec.
I hope, this will help you
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