A bunch of fake Indians are on a ship docked in Boston Harbor and want to push a
ID: 1324373 • Letter: A
Question
A bunch of fake Indians are on a ship docked in Boston Harbor and want to push all the crates of tea over the edge of the ship, into the water. They set up a ramp to do It. The ramp is 10.0m long, and its end is propped up so that it is 5.00 m above the deck. The deck of the ship is 15.0 m above the surface of the water. Each crate of tea has a mass of 50.0 kg. They push the crates of tea all the way up the 10.0 m ramp two at a time, starting from the bottom. (Ignore any friction.) a.If the Indians push with a combined force, F = 1,100 N parallel to the plank, What is the horizontal distance from the edge of the plank, d, that the crates hit The water? b.On one of their runs up the ramp, the top crate falls off half way up the ramp. The Indians continue to push with the same combined force of F = 1,100 N parallel to the plank. what is the horizontal distance from the edge of the Plank, d, that the single remaining crate now hits the water?Explanation / Answer
angle of the deck @ =tan-1(5/10) = 26.57 degree
using work -energy theorem,
work done by gravity + work done by worker = change in K.E.
- 100 x 9.81 x 5 + 1100 x10 = 100v^2 /2 - 0
v =11.04 m/s
now in vertical using h = ut + at^2 /2
-(5 + 15) = 11.04sin26.57 *t - 9.81t^2 /2
4.9t^2 - 4.94t -20 = 0
solving t = 2.59 sec
d = ux * t
d = 11.04cos26.57 * 2.59 = 25.57 m
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using work -energy theorem for half the distance
work done by gravity + work done by worker = change in K.E.
-100 x 9.81 x5sin26.57 + 1100 x 5 = 100u^2 /2 - 0
u = 8.13 m/s
for further half distance,
-50 x 9.81 x (5 - 5sin26.57) + 1100 x 5 = 50v^2 /2 - 50*8.13^2 /2
v = 15.23 m/s
now using
h = ut + at^2 /2 in vertical
-20 = 15.23sin26.57 -9.81t^2 /2
4.9t^2 - 6.81 -20 = 0
t = 2.83 sec
d = ux * t = 15.23 cos26.57 *2.83 =38.55 m
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