A child slides a block of mass 2 kg along a slick kitchen floor. If the initial speed is 4m/s and the block hits a spring with spring constant 6 N/m, What is the result if the block slides across 2m of a rough floor that ha μk = 0.2?
getting to the spring..
vf2=v02+2as
F(friction)=μ(mg) ------------> a=μg
vf2=v02+2(μg)s
vf2= 23.848
at the spring..
vf2=v02+2as
vf2=0
v02=23.848
[itex]\Sigma[/itex]F=F(friction)+F(spring)=-m|μg|-m|kx|
a=-|μg|-|kx|
v02=2(-|μg|-|kx|)s
but when i plug all the numbers in.. i get a quadratic equation and it ends up having an imaginary component because the "b^2" term, within the quadratic formula ends up being less than the "4ac" term. i COULD go through and just solve it.. but i don't know what to do because i don't know what to do with the imaginaries.
getting to the spring..
vf2=v02+2as
F(friction)=μ(mg) ------------> a=μg
vf2=v02+2(μg)s
vf2= 23.848
at the spring..
vf2=v02+2as
vf2=0
v02=23.848
[itex]\Sigma[/itex]F=F(friction)+F(spring)=-m|μg|-m|kx|
a=-|μg|-|kx|
v02=2(-|μg|-|kx|)s
but when i plug all the numbers in.. i get a quadratic equation and it ends up having an imaginary component because the "b^2" term, within the quadratic formula ends up being less than the "4ac" term. i COULD go through and just solve it.. but i don't know what to do because i don't know what to do with the imaginaries.
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