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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 maximum compression of the spring? What is the result if the block slides across 2m of a rough floor that ha μk = 0.2?

(second part of the problem)

getting to the spring..

v

F(friction)=μ(mg) ------------> a=μg

v

v

at the spring..

v

v

v

[itex]\Sigma[/itex]F=F(friction)+F(spring)=-m|μg|-|kx|

a=-|μg|-(|kx|/m)

v

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.

(second part of the problem)

getting to the spring..

v

_{f}^{2}=v_{0}^{2}+2asF(friction)=μ(mg) ------------> a=μg

v

_{f}^{2}=v_{0}^{2}+2(μg)sv

_{f}^{2}= 8.152m/sat the spring..

v

_{f}^{2}=v_{0}^{2}+2asv

_{f}^{2}=0v

_{0}^{2}=8.152[itex]\Sigma[/itex]F=F(friction)+F(spring)=-m|μg|-|kx|

a=-|μg|-(|kx|/m)

v

_{0}^{2}=2(-|μg|-(|kx|/m))sbut 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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