In the figure, a block slides along a track from one level to a higher level after passing through an intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional force stops the block in a distance d. The blocks initial speed v0 is 6.0m/s, the height difference h is 1.1m, and μk is 0.60. Find d.
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I know that I must use the change in Mechanical Energy (Change in Kinetic + Change in Potential), and must also incorporate the change in Thermal Energy due to heat, but I'm not sure how to go about it because I'm not getting the correct answer, which was solved out to be 1.2 meters. Any help would be greatly appreciated!
http://s3.amazonaws.com/answer-board-image/3da740ab-48ed-4fd6-94d0-e76dcefd1da3.gif <--Image
I know that I must use the change in Mechanical Energy (Change in Kinetic + Change in Potential), and must also incorporate the change in Thermal Energy due to heat, but I'm not sure how to go about it because I'm not getting the correct answer, which was solved out to be 1.2 meters. Any help would be greatly appreciated!
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Set h = 0 where the block starts. Then determine the speed of the block at the hight h:
E tot = E ' tot
KE1 = PE g + KE2
(1/2)m(Vo)^2 = mgh + (1/2)m(v2)^2
(Vo)^2 = 2gh + (v2)^2
v2 = sqrt[(Vo)^2 - 2gh]
v2 = sqrt[(6.0 m/s)^2 - 2(9.81 m/s^2)(1.1 m)]
v2 = 3.797 m/s
Now use the work-energy theorem:
Δ Ke = W(Fk)
KE1 - KE2 = (Fk)(d)
But we know that KE2 = 0 J because the block stops (so it's velocity is zero):
KE1 = (Fk)(d)
d = [KE1] / [Fk]
d = [KE1] / [μkFn]
But Fn = Fg = mg from a simple free body diagram
d = [KE1] / [μkmg]
d = [(1/2)(m)(v2)^2] / [(μk)(m)(g)]
d = [(1/2)(v2)^2] / [(μk)(g)]
d = [(1/2)(3.797 m/s)^2] / [(0.60)(9.81 m/s^2)]
d = 1.22 m
Done!
E tot = E ' tot
KE1 = PE g + KE2
(1/2)m(Vo)^2 = mgh + (1/2)m(v2)^2
(Vo)^2 = 2gh + (v2)^2
v2 = sqrt[(Vo)^2 - 2gh]
v2 = sqrt[(6.0 m/s)^2 - 2(9.81 m/s^2)(1.1 m)]
v2 = 3.797 m/s
Now use the work-energy theorem:
Δ Ke = W(Fk)
KE1 - KE2 = (Fk)(d)
But we know that KE2 = 0 J because the block stops (so it's velocity is zero):
KE1 = (Fk)(d)
d = [KE1] / [Fk]
d = [KE1] / [μkFn]
But Fn = Fg = mg from a simple free body diagram
d = [KE1] / [μkmg]
d = [(1/2)(m)(v2)^2] / [(μk)(m)(g)]
d = [(1/2)(v2)^2] / [(μk)(g)]
d = [(1/2)(3.797 m/s)^2] / [(0.60)(9.81 m/s^2)]
d = 1.22 m
Done!
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The block has initial KE = 1/2m(v0)² = (0.5)m(6)² = 18m
If we measure the GPE of the block from the initial level, then it = 0
TOTAL ME initially = KE + GPE = 18m
The block loses no ME traversing the path in blue where µk = 0
When the block STARTS traversing "d", its KE = 1/2mv²
(0.5)mv² = 18m - GPE = 18m - mgh = 18m - m(9.8)(1.1) = 18m - 10.78m = 7.22m
v² = 7.22/0.5 = 14.44
v = 3.8 m/s <= speed of block at start of its travel of displacement = d
friction force = µk(mg) = (0.60)m(9.8) = 5.88m
deceleration caused by friction force = 5.88m/m = 5.88 m/s²
time to stop = t = v/5.88 = 3.8/5.88 = 0.646 s
d = vt - 1/2(5.88)t² = (3.8)(0.646) - (0.5)(5.88)(0.646)² = 2.4548 - 1.2269 = 1.2 m ANS
If we measure the GPE of the block from the initial level, then it = 0
TOTAL ME initially = KE + GPE = 18m
The block loses no ME traversing the path in blue where µk = 0
When the block STARTS traversing "d", its KE = 1/2mv²
(0.5)mv² = 18m - GPE = 18m - mgh = 18m - m(9.8)(1.1) = 18m - 10.78m = 7.22m
v² = 7.22/0.5 = 14.44
v = 3.8 m/s <= speed of block at start of its travel of displacement = d
friction force = µk(mg) = (0.60)m(9.8) = 5.88m
deceleration caused by friction force = 5.88m/m = 5.88 m/s²
time to stop = t = v/5.88 = 3.8/5.88 = 0.646 s
d = vt - 1/2(5.88)t² = (3.8)(0.646) - (0.5)(5.88)(0.646)² = 2.4548 - 1.2269 = 1.2 m ANS