Problem 3/202

If anyone is still having trouble with this question here is a little tutorial to help. I found that there are two different but essentially the same methods of doing this problem. I guess its just a matter of which is easier to see. I see the first way easier. Please post if you find a mistake.

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--Method 1--

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This method solves for the results of the blocks state at the time when P changes magnitudes using conservation of momentum. It then uses that state of the block to find the time time till it stops.

F_friction = (mg)*(u_k) = (20)(0.2) = 4 lbs

P= 5     0 to 0.2

P=2.5    t above 0.2

Using conservation of momentum:

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Interval  0 to 0.2

F_net_1 = P + F_friction = 5 + 4 = 9 lbs

EQ1: (m)(V_0) + Integral(F_net_1 dt) from 0 to 0.2 = m(V_2)

"V_0 is the initial velocity."

"V_2 would be the velocity of the block after the first time interval."

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Interval  t above 0.2

F_net_2 = P+F_friction = 6.5 lbs

EQ2: m(V_2) + Integral(F_net_2 dt) from 0.2 to t = m(V_f)

"V_f is the final velocity of the block."

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Now we combine the equations for an equation with only t as the unknown.

EQ_net = EQ1 +EQ2

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--Method 2--

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This is the method that was shown in the lecture video. Instead of reflecting how the block's state is in relationship to time it instead reflects how each force is playing a role in bring the block to rest.

F_friction = (mg)(u_k) = (20)(0.2) = 4lbs

Using conservation of momentum:

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We know that the force P changes at t=0.2s so we will have two separate integrals to explain it's contribution.

P_1 = 5 lbs

P_2 = 2.5 lbs

P_momentum =

    Integral(P_1 dt) from 0 to 0.2 } first interval

+ Integral(P_2 dt) from 0.2 to t } second interval

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Now the friction's contribution. 

F_momentum = Integral(F_friction dt) from 0 to t

t is for the whole interval since kinetic friction acts till stop.

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These equations then combine to find the final velocity. This results in an equation with only t as the variable.

P_momentum + F_momentum - (m)(V_f) = 0