F=18-0.53•x
a=F/m=(18-0.53•x)/5.5
v=sqrt(2•a•x)=sqrt[(18-0.53•x)x/5.5]=
= sqrt[(18 - 0.53•12) •12/5.5=5.04 m/s
force in the x-direction with magnitude Fx=18N-(0.530N/m)x is applied to a 5.50 box that is sitting on the horizontal, frictionless surface of a frozen lake. is the only horizontal force on the box.
what is the speed after 12m. starting from rest and x=0
4 answers
the first answer is wrong
Integral of F(x) = .5mv^2
18x-.5(.53)x^2=.5mv^2
find v.
18x-.5(.53)x^2=.5mv^2
find v.
8.04m/s