A boulder is raised above the ground, so that its potential energy relative to the ground is 200 J. Then it is dropped. Estimate
1 answer:
Answer:
200 J
Explanation:
In this problem, I assume there is no air resistance, so the system is isolated (=no external forces).
For an isolated system, the total mechanical energy is constant, and it is given by:
where
KE is the kinetic energy
PE is the potential energy
The kinetic energy is the energy due to the motion of the object, while the potential energy is the energy due to the position of the object relative to the ground.
At the beginning, when the boulder is raised above the ground, its height above the ground is maximum, while its speed is zero; it means that all its mechanical energy is just potential energy, and it is:
As the boulder falls down, its altitude decreases, so its potential energy decreases, while the speed increases, and the kinetic energy increases. Therefore, potential energy is converted into kinetic energy.
Eventually, just before the boulder hits the ground, the height of the object is zero, and the speed is maximum; this means that all the energy has now converted into kinetic energy, and we have
Therefore, the kinetic energy just before hitting the ground is 200 J.
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Answer:
r = <2.640 10⁶, 1.01 10⁸, 0> m
Explanation:
For this exercise we are going to solve it for each direction separately,
we locate a fixed reference frame in space at the height of the rocket, such that the position of the rocket is
r₀ = <4, 4, 0> 10³ m
X axis
the initial velocity of the ship on this axis is v₀ₓ = 0, when it passes through the point x₀ = 4 km it ignites the rockets, experiencing a force of Fₓ = 7.0 10⁵ N for 21 s and the rockets turn off
They ask where it is after one hour t = 1 h = 3600 s
Let's apply Newton's second law
Fₓ = m aₓ
aₓ = Fₓ / m
aₓ = 7.0 10⁵/2 10⁴
aₓ = 3.5 10¹ m / s
Let's use kinematics to find the distance
for the first t₁ = 21 s the movement is accelerated
x₁ = x₀ + v₀ t₁ + ½ aₓ t₁²
x₁ = x₀ + ½ aₓ t₁²
x₁ = 4000 + ½ 35 21² = 4000 + 7717,.5
x₁ = 11717.5 m
this instant has a speed of
vₓ = v₀ₓ + aₓ t
vₓ = aₓ t ₁
vₓ = 35 21
vₓ = 735 m / s
the rest of the time there is no acceleration so it is uniform motion at this speed
t₂ = 3600 - 21
t₂ = 3576 s
vₓ = x₂ / t₂
x₂ = vₓ t₂
x₂ = 735 3576
x₂ = 2,628 10⁶ m
the total distance traveled in this direction is
x_total = x₁ + x₂
x_total = 11717.5 + 2.628 10⁶
x_total = 2,640 10⁶ m
Y axis
on this axis it is in the initial position of y₀ = 4000 m, with an initial velocity of = 28 10³ m / s and there is no force on this axis F_{y} = 0
The movement in this axis is uniform,
v_{y} = y / t
y = v_{y} t
y = 28 10³/3600
y = 1.01 10⁸ m
the total distance is
y_total = y₀ + y
y_total = 4000 + 1.01 10⁸
y_total = 1.01 10⁸ m
Z axis
the initial position is z₀ = 0, with an initial velocity of v₀ = 0 and in this axis there is no force F_{z} = 0
the movement is uniform
z = 0
the final position of the rocket after 1 h is
r = <2.640 10⁶, 1.01 10⁸, 0> m