After a tornado, a 0.55g straw was found embedded 2.3cm into the trunk of a tree. if the average force excerted on the straw by the tree was 65N, what was the speed of the straw when it hit the tree?
(Kinetic energy of straw) = Work = (Force)x(penetration distance)
(1/2) M V^2 = F*X
V = sqrt(2F*X/M)
2.32
74 m/s
To find the speed of the straw when it hit the tree, we can use the principle of conservation of energy. The energy of the moving straw transforms into potential energy when it comes to rest inside the tree trunk.
First, we need to find the potential energy gained by the straw when it is embedded into the tree trunk. We can use the formula:
Potential Energy (PE) = Force × Distance
The force exerted on the straw by the tree is given as 65N, and the distance the straw is embedded into the tree trunk is 2.3cm, which is equivalent to 0.023m. Therefore,
PE = 65N × 0.023m
PE = 1.495 Joules
Next, we can equate the potential energy gained by the straw with the initial kinetic energy it possessed before hitting the tree. The kinetic energy formula is:
Kinetic Energy (KE) = 0.5 × Mass × Velocity^2
The given mass of the straw is 0.55g, which is equivalent to 0.00055kg. Let's assume the initial velocity of the straw is v.
KE = 0.5 × 0.00055kg × v^2
Since the straw comes to rest inside the tree trunk, its final velocity is 0. Therefore, the equation becomes:
1.495 J = 0.5 × 0.00055kg × 0^2
Simplifying the equation, we find that the speed of the straw when it hit the tree is approximately 27.52 m/s.
The speed of the straw when it hit the tree is approximately 27.52 m/s.