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

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

An object with a mass of 4 kg4 kg is traveling at 3 m/s3 m/s. If the object is accelerated by a force of f(x) = 2x^2 -x +3 f(x) = 2x^2 -x +3 over x in [0, 9]x in [0, 9], where x is in meters, what is the impulse at x = 2x = 2?