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"Two masses comprise an inertial system: the first, a 2 Kg mass which rests on a rough table top; then second, 5 Kg mass which hangs freely 0.4 m above the floor from an ideal pulley at the edge of the table top. The two masses are connected by an inextensible string run over the pulley. The coefficient of kinetic friction between the table top and the first mass is .40."

first I tried

[tex]m_{1} = 2 kg[/tex]

[tex]T-F_{k}=m_{1}a \ where\ F_{k}= \mu N \ and \ N = m_{1}g[/tex]

so

[tex]T=m_{1}a+ \mu m{1}g[/tex]

Then mass 2

[tex] m_{2} = 5 kg[/tex]

[tex] F= MA [/tex]

[tex]m_{2}g -T = m_{2}a[/tex]

substituting T for what T = in part 1 gives me

[tex]m_{2}g -m_{1}a - \mu m{1}g = m_{2}a[/tex]

so a = [tex] (m_{2}g - \mu m_{1}g)/(m_{1} + m_{2}) [/tex]

pluging in 5 kg for m2 98. m/s^2 for g .4 for mu and 2 kg for m1 gives me an acceleration of5.88 m/S^2

then to solve for velocity I would go [tex] W_{nc} = E_{f} - E_{i} [/tex]

so [tex] -\mu m_{1}gd = 1/2m_{1}v^2 + 1/2 m_{2}v^2-m_{2}gd[/tex]

then solving using the same numbers above and using .4 for d v comes out to be2.55 m/s

could I also solve for T in equation 1 to get a conservative work force of 19.6 J?

someone also showed me that

deltaK = work total = work gravity + work friction

which is 1/2(m2 + m1)(v^2 -0) = -m2g(-d -0) + (- mu m1g)d

then it comes down to v = [tex] \sqrt{\frac{2gd(m_{2}- \mu m_{1}}{m_{2}+m_{1}}} [/tex]

but that gives me a v = 4.704 m/s which is not 2.55 m/s

and someone told me the anwser to the velocity component is 2.88 m/s

I am confused

moreover, how could i figure out how far it traveled once the free hanging block hit the ground?

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# Homework Help: Help with simple work / nrg conservation question

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