A constant force acts for $5$ seconds on abody of mass $5$ quitals and then act.During the next $5$ secs the body describe $100$ m. Find the magnitude of the force

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Solution

we know that 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/55²)
x=40/3
and accn=x/5
=40/35
=8/3
and F= ma
=1008/3=266.66N

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A constant force acts for 5 seconds on abody of mass 5 quitals and then act.During the next 5 secs the body describe 100m. Find the magnitude of the force

we know that by Newton's 2nd lawF=m(x-u)/t=m(x-0)/t=mx/5a=F/m=x/5where x is its final velocity after 5 sec.Also by usingv=u+atv=x+(x/5)*5=2xBy usings=ut+1/2(at²)we get100=5x+1/2(x/5*5²)x=40/3and accn=x/5=40/3*5=8/3and F= ma=100*8/3=266.66N

A constant force acts for $5$ seconds on abody of mass $5$ quitals and then act.During the next $5$ secs the body describe $100$ m. Find the magnitude of the force

we know that 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/55²)
x=40/3
and accn=x/5
=40/35
=8/3
and F= ma
=1008/3=266.66N