The spring has been stretched 0.701 m
Explanation:
The elastic potential energy of a spring is the potential energy stored in the spring due to its compression/stretching. It is calculated as
where
k is the spring constant
x is the elongation of the spring with respect to its equilibrium position
For the spring in this problem, we have:
E = 84.08 J (potential energy)
k = 342.25 N/m (spring constant)
Therefore, its elongation is:
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Answer:
Explanation:
<u>Motion in The Plane</u>
When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.
The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of and
as the initial speed and angle, then we have
If we want to know the maximum height reached by the object, we find the value of t when becomes zero, because the object stops going up and starts going down
Solving for t
Then we replace that value into y, to find the maximum height
Operating and simplifying
We have
The maximum height is
