The same bird on the tree has more gravitational potential energy. This is because it is at a higher distance from the ground as it is on the tree, than when it is on the ground.
Considering also the formula for Gravitational Potential Energy GPE = mgh
For the bird on the ground, h =0, therefore GPE = m*9.8*0 = 0
For that on the tree = mgh = m*9.8*h
Of course the one on the tree has a value greater than zero.
Let <em>F₁ </em>and <em>F₂</em> denote the two forces, and <em>R</em> the resultant force.
<em>F₁ </em>and <em>F₂</em> point perpendicularly to one another, so their dot product is
<em>F₁ </em>• <em>F₂</em> = 0
<em />
We're given that one of these vectors, say <em>F₁</em>, makes an angle with <em>R</em> of 30°, so that
<em>F₁</em> • <em>R</em> = ||<em>F₁</em>|| ||<em>R</em>|| cos(30°)
But we also have
<em>F₁</em> • <em>R</em> = <em>F₁ </em>• (<em>F₁ </em>+ <em>F₂</em>) = (<em>F₁ </em>• <em>F₁</em>) + (<em>F₁ </em>• <em>F₂</em>) = <em>F₁ </em>• <em>F₁ </em>=<em> </em>||<em>F₁</em>||²
So, knowing that ||<em>R</em>|| = 100 N, we get that
(100 N) ||<em>F₁</em>|| cos(30°) = ||<em>F₁</em>||²
(100 N) cos(30°) = ||<em>F₁</em>||
||<em>F₁</em>|| ≈ 86.6 N
(And the same would be true for <em>F₂</em>.)
Explanation:
speed of curved path=under root of Mew and gravity
smooth surface Mew=0
so under root of 0×g
=0
