The force applied against an object moving horizontally on a linear path is described by #F(x)=x^2-3x + 3 #. By how much does the object's kinetic energy change as the object moves from # x in [ 0 , 1 ]#?

1 Answer
Apr 13, 2016

Answer:

Newton's second law of motion:

#F=m*a#

Definitions of acceleration and velocity:

#a=(du)/dt#

#u=(dx)/dt#

Kinetic energy:

#K=m*u^2/2#

Answer is:

#ΔK=11/6# #kg*m^2/s^2#

Explanation:

Newton's second law of motion:

#F=m*a#

#x^2-3x+3=m*a#

Substituting #a=(du)/dt# does not help with the equation, since #F# isn't given as a function of #t# but as a function of #x# However:

#a=(du)/dt=(du)/dt*(dx)/dx=(dx)/dt*(du)/dx#

But #(dx)/dt=u# so:

#a=(dx)/dt*(du)/dx=u*(du)/dx#

Substituting into the equation we have, we have a differential equation:

#x^2-3x+3=m*u(du)/dx#

#(x^2-3x+3)dx=m*udu#

#int_(x_1)^(x_2)(x^2-3x+3)dx=int_(u_1)^(u_2)m*udu#

The two speeds are unknown but the positions #x# are known. Also, mass is constant:

#int_(0)^(1)(x^2-3x+3)dx=m*int_(u_1)^(u_2)udu#

#[x^3/3-3x^2/2+3x]_0^1=m*[u^2/2]_(u_1)^(u_2)#

#(1^3/3-3*1^2/2+3*1)-(0^3/3-3*0^2/2+3*0)=m*(u_2^2/2-u_1^2/2)#

#11/6=m*u_2^2/2-m*u_2^2/2#

But #K=m*u^2/2#

#11/6=K_2-K_1#

#ΔK=11/6# #kg*m^2/s^2#

Note : the units are #kg*m^2/s^2# only if the distances given #(x in[0,1])# are in metres.