The drawing shows two frictionless inclines that begin at theground level (h = 0

Question

The drawing shows two frictionless inclines that begin at theground level (h = 0 m) and slope upward at the same angle . Onetrack is longer than the other, however. Identical blocks areprojected up each track with the same initial speed vo. On thelonger track the block slides upward until it reaches a maximumheight H above the ground. On the shorter track the block slidesupward, flies off the end of the track at a height H1 above theground, and then follows the familiar parabolic trajectory ofprojectile motion. At the highest point of this trajectory, theblock is at a height H2 above the end of the track. The totalmechanical energy of each block is the same (Why?) everywhere,although the mix of kinetic and potential energies will bedifferent for the blocks depending on where the blocks are.

Suppose the initial speed of each block is vo = 6.45 m/s, and eachincline slopes upward at an angle of = 51.1°. The block on theshorter track leaves the track at a height of H1 = 1.44 m above theground.

(1) What is the value of H for the block on the longer track?
H =_____________

(2) What is the total height H1 + H2 for the block on the shortertrack?
H1 + H2 =______________ Please show work

Explanation / Answer

Since the only force acting on the blocks is a conservativeforce, that of gravity, the total mechanical energy of the blocksis constant, by definition. Therefore, for the first block,the total energy that it has at the bottom will be equal to thetotal energy it has at the peak of its climb. Initially, allof the energy of the block is in kinetic energy, so: Eo = Ko + Uo = Ko= mvo2/2 At the top of the climb, all of its energy is in potentialenergy, so we have that: Ef = Kf + Uf = Uf= mgH Since energy is conserved, Eo = Ef, somvo2/2 = mgH, and thus: H = vo2/2g = (6.45m/s)2/2(9.8 m/s2) = 2.12m On the second block, when it leaves the ramp, it will haveboth potential and kinetic energy. We have that: El = Kl + Ul =mv2/2 + mgH1 Since energy is conserved, El = Eo, sowe have that: mv2/2 + mgH1 =mvo2/2 v2 = vo2 -2gH1 = (6.45 m/s)2 - 2(9.8m/s2)(1.44 m) = 13.38m2/s2 v = 3.66 m/s This is at an angle that is 51.1o from thehorizontal, so the horizontal speed of the block as it leaves theramp is going to be equal to: vh = vcos = (3.66m/s)cos(51.1o) = 2.30 m/s Since the block will not accelerate horizontally when it isflying through the air, but at its peak the vertical velocity willbe zero, this will be the velocity of the block at its peak. The total energy of the block at its peak is equal to: Ep = Kp + Up =mvh2/2 + mgH2 Since energy is conserved, this is equal to the initial energyof the block, so we have that: mvh2/2 + mgH2 =mvo2/2 H2 = (vo2 -vh2)/2g = ((6.45 m/s)2 -(2.30 m/s)2)/2(9.8 m/s2) = 1.85 m Therefore, H1 + H2 = 1.44 m +1.85 m =3.29 m Since energy is conserved, this is equal to the initial energyof the block, so we have that: mvh2/2 + mgH2 =mvo2/2 H2 = (vo2 -vh2)/2g = ((6.45 m/s)2 -(2.30 m/s)2)/2(9.8 m/s2) = 1.85 m Therefore, H1 + H2 = 1.44 m +1.85 m =3.29 m

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