A constant force acts for 5 second a body of mass 5 quitals and then ceased to act. during next 5 seconds the body describes 100m . find the magnitude of force.
Answers
Answered by
33
By Newton's 2nd law
F=m(x-u)/t
=m(x-0)/t
=mx/5
a=F/m=x/5
where x is its final velocity after 5 sec.
Also by using
v=u+at
v=x+(x/5)*5
=2x
By using
s=ut+1/2(at²)
we get
100=5x+1/2(x/5*5²)
x=40/3
and accn=x/5
=40/3*5
=8/3
and F= ma
=100*8/3=266.66N
karthik202:
Not the correct answer
Oops i took the mass of body 1 quintol instead of 5.In the last step multiply by 5 and you get your answer.
Answered by
91
v = u + at
v = 0 + a x 5 (u = 0)
v = 5 a
after 5 seconds
distance s = v x t = 5 a x 5
100 = 25 a => a = 4 m/s^2
F = ma = 500 kg x 4 m/s^2 (1 quintal = 100 kg => 5 quintals = 500 kg)
F = 2000 N
hope this will help you.
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