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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

Leo Abeyta

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#

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Answer 2

Axel Acevedo

Impulse at x = 8 is approximately 400 Ns.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.