Question

An object with a mass of `2 kg` is traveling at `14 m/s`. If the object is accelerated by a force of `f(x) = x+xe^x` over `x in [0, 9]`, where x is in meters, what is the impulse at `x = 8`?

Answer 1

1

Explanation:

The definition of impulse is

`I=F*Deltat =`

and using the 2nd law of Newton `F=m*a` it is equivalent to the variation of momentum,

`I=Deltap = m*Deltav = m*(v_f -v_0) ; m=2kg ; v_0=14m/s`

so, we need to calculate the speed at the position `x=8; v_f`

By the 2nd law of Newton `F=m*a => F(x) = m* (dv)/dt`

`F(x) = m*(dv(x))/dx * (dx)/dt = m *(dv(x))/dx *v(x)`

`int_0^8F(x)dx = m*(v_f^2 -v_0^2)`

`v_f = sqrt (v_0^2+1/m*int_0^8F(x)dx)`

thus, the impulse is

`I=m*(sqrt (v_0^2+1/m*int_0^8F(x)dx) -v_0) = 262.5* kg* m/s`