General Physics Indians push 100 kg crates up a 10.0 m plank with a force, F, pa
ID: 1283495 • Letter: G
Question
General Physics
Indians push 100 kg crates up a 10.0 m plank with a force, F, parallel to the plank.
The right side of the plank is 5.00 m above the deck of the ship, and 20.0 m above the water
surface.
Please explain thoroughly. Thank you!
4) 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 10.0 m 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
a) The work performed on the crates is calculated as the product of force and displacement. The force applied by the Indians is 1100. However, there is also a negative force by gravity acting downward = m X 9.8 = 100 X 9.8 = 980N
The component of this force in the direction opposite to the movement of the crates = 980 cos (60) = 490N
Thus the net force acting on the crates = 1100-490 = 610N
The total displacement energy transferred to the crates would be converted to kinetic energy once they leave the ramp. Thus Force X Displacement = 610 X10 = 0.5 X Mass X Velocity2 = 0.5 X 100 X Velocity2
Thus 122 = Velocity2 hence leaving velocity = 11.045 m/s at an angle of 30 degrees to the horizontal.
The horizontal component of this Velocity = Vx=11.045 X cos (30) = 9.56 m/s
The Vertical component Vy=11.045 X sin (30) = 5.52 m/s
Use the vertical component to determine the time take to hit the water. Multiply this time by the horizontal component to obtain the distance at which it hits the surface.
The time taken to hit the water can be broken down into : Time taken to reach the highest point and back to deck level + time taken to hit the surface from deck level.
The first part is calculated using the formula v=u + at at the highest point, v=0; u=5.52, a=g=-9.8
Hence 5.52= 9.8 *t hence t=0.563 hence total time taken = 2*t=1.127 seconds
The second part is calculated using the formula s= ut + 0.5 at2
Here s=20, u=-5.52, a=g=-9.8 hence we have the equation 4.9t2+5.52t+ 20=0
Hence we have t=1.534 seconds adding both parts we have
Total time taken to hit water surface = 1.127 + 1.534 = 2.66 seconds
Multiplying this with the horizontal component of velocity i.e 9.56 m/s we have the distance from the ship at which the crates hit the surface as 9.56 X 2.66= 25.43 meters
b) The work performed on the crates is calculated as the product of force and displacement. The force applied by the Indians is 1100. However, there is also a negative force by gravity acting downward = m X 9.8 = 100 X 9.8 = 980N
The component of this force in the direction opposite to the movement of the crates = 980 cos (60) = 490N
Thus the net force acting on the crates = 1100-490 = 610N
The total displacement energy transferred to the crates would be converted to kinetic energy once they leave the ramp. Upto the halfway mark the energy transferred to both crates is 610X5 =3050 units of work
At the halfway mark one of the crates falls of and hence the resisting gravitational force is now 245N. Hence the net force acting on the single remaining crate = 1100-245= 855N The work done is calculated as 855X5=4275 units of work. Thus total energy transfered = 3050+4275=7325 units of work
Thus 7325 = 0.5 X Mass X Velocity2 = 0.5 X 50 X Velocity2
Thus 293 = Velocity2 hence leaving velocity = 17.11 m/s at an angle of 30 degrees to the horizontal.
The horizontal component of this Velocity = Vx=17.11 X cos (30) = 14.817 m/s
The Vertical component Vy=17.11 X sin (30) = 8.55 m/s
Use the vertical component to determine the time take to hit the water. Multiply this time by the horizontal component to obtain the distance at which it hits the surface.
The time taken to hit the water can be broken down into : Time taken to reach the highest point and back to deck level + time taken to hit the surface from deck level.
The first part is calculated using the formula v=u + at at the highest point, v=0; u=8.55, a=g=-9.8
Hence 8.55= 9.8 *t hence t=0.873 hence total time taken = 2*t=1.746 seconds
The second part is calculated using the formula s= ut + 0.5 at2
Here s=20, u=-8.55, a=g=-9.8 hence we have the equation 4.9t2+8.55t+ 20=0
Hence we have t=1.328 seconds adding both parts we have
Total time taken to hit water surface = 1.746 + 1.328 = 3.074 seconds
Multiplying this with the horizontal component of velocity i.e 14.817 m/s we have the distance from the ship at which the crates hit the surface as 14.817 X 3.074= 45.55 meters
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