Look at this problem. Your friend\'s Frisbee has become stuck 16m above the grou

Question

Look at this problem.
Your friend's Frisbee has become stuck 16m above the ground in a tree. You want to dislodge the Frisbee, by throwing a rock at it. The Frisbee is stuck pretty tight, so you figure that the rock needs to be traveling at least 4.0 m/s wen it hits the Frisbee. If you release the rock at 2.0 m above the ground, with with minimum speed must you throw it?

Few Obvious possibilities for each:

System – only the rock or the rock and the Earth
Zero of potential energy – ground level, the height where the rock is released, or the height of the frisbee

These questions will look at the consequences if you use a different choice for the zero of the potential energy.

(a) Which point did you choose for the zero of the potential energy in that problem? What are the initial and final heights using this choice? Write the conservation of energy equation.

(b) Select one of the other possibilities for the zero of potential energy listed above. This redefines the origin of your coordinate system. What are the initial and final heights using this choice? Rewrite the conservation of energy equation using this new coordinate system.

(c) Demonstrate that the two conservation of energy equations in parts a and b give the same answer for the minimum speed necessary for the rock to dislodge the frisbee.

Explanation / Answer

applying Conservation of energy we have
note i chose 2m above ground level as potential zero zone
letss say rock's initial velocity is v initial height =0 (2m is taken as zero) and final height is 14m (16-2)
we have
K.E+P.E = constant
so
0+0.5*mv^2=mg*14+0.5*m*4^2
==>v^2=290.4
==>v=17.04m/sec  <=====part a

suppose we choose ground as zero reference (initial h=2m final h=16m)

we have

K.E+P.E = constant
so

mg*2+0.5*mv^2=mg*16+0.5*m*4^2

==>0.5*mv^2=mg*14+0.5*m*4^2

==>v^2=290.4
==>v=17.04m/sec <=====part b
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