An object with a mass of #4 kg# is traveling at #1 m/s#. If the object is accelerated by a force of #f(x) = x^2 -x +1 # over #x in [1, 9]#, where x is in meters, what is the impulse at #x = 2#?

1 Answer
Feb 8, 2016

Answer:

Impulse #J = \Delta p = 1.5377 N.s#

Explanation:

Impulse-Momentum Theorem: #\vec{J_{}}=\Delta\vec{p_{}}#,

To calculate the Impulse ( #\vec{J}#) just calculate the change in momentum ( #\Delta\vec{p_{}}# ). To calculate the change in momentum, remember the relation between momentum (#p#) and the kinetic energy (#K#) :
#K=p^2/(2m) \rightarrow p=\sqrt(2mK)#
Therefore #\Delta p = \sqrt(2m)(\sqrt(K_f)-\sqrt(K_i))#

Work-Energy Theorem: The total work done by all the forces acting on the object must be equal to the change in its kinetic energy #W=\Delta K = K_f-K_i#, where #K_i# and #K_f# are the initial and final kinetic energies.
#K_f = K_i + W#

#K_i = 1/2 mv_i^2 = 1/2(4kg)(1m/s)^2=2# Joules,
#W=\int_1^2F(x)dx=\int_1^2(x^2-x+1)dx=[x^3/3-x^2/2+x]_1^2#
#\qquad\quad##=(2^3/3-2^2/2+2)-(1/3-1/2+1)=11/6# Joules
#K_f = K_i + W = 2+11/6=23/6# Joules

Now we are ready to evaluate the impulse:
#J=\Delta p = \sqrt(2m)(\sqrt(K_f)-\sqrt(K_i)) = 1.5377 N.s#