An object with a mass of #4 kg# is traveling at #3 m/s#. If the object is accelerated by a force of #f(x) = 2x^2 -x +3 # over #x in [0, 9]#, where x is in meters, what is the impulse at #x = 2#?

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Answer 1 · 2016-06-30T08:46:23

#3.04(kgm)/s#

Given

#m-"Mass of the oject"=4kg#

#u->"Initial velocity"=3m/s#

#"Force ",f(x)=2x^2-x+3#

So #"Acceleration " a(x)=1/m*f(x)#

Again we know #a(x)=v(dv)/(dx)#

So
# v(dv)/(dx)=1/mf(x)=1/4(2x^2_x+3)#

#=>intvdv=1/4int(2x^2-x+3)dx#

#=>v^2/2=1/4(2*x^3/3-x^2/2+3x)+c#

#=>v^2=x^3/3-x^2/4+3/2x+c#

#"Given at " x=0, v=3#

we get c=9

So now

#v(x)=sqrt(x^3/3-x^2/4+3/2x+9)#

When x = 2 velocity becomes

#v(2)=sqrt(2^3/3-2^2/4+3/2*2+9)#

#=sqrt(8/3-1/2+3+9)=sqrt(14.16)m/s=3.76m/s#

Hence momentum of the object at x=2 is

#"Momentum "= "mass"xx"velocity"#
#=4xx3.76=15.04"kgm"/s#

So impulse which is change in momentum during the time to reach at #x=2# from #t =0# will be

#I=(15.04-3*4)"kgm/"s=3.04kgm"/"s#