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myrzilka [38]
11 months ago
11

A ball is thrown vertically upward. As it rises, what happens to its total energy?.

Physics
1 answer:
diamong [38]11 months ago
3 0

The total energy of a ball stays constant as it is thrown upward because potential energy increases while kinetic energy decreases. When the ball reaches its maximum height, the velocity is zero. Therefore, only potential energy exists rather than kinetic energy.

The thrower's movement imparts kinetic energy to a ball thrown vertically. The maximum height that can be achieved after leaving the hand will depend on the actual velocity. Air resistance causes some of this energy to be lost to the air as frictional dissipation, which warms the air in the area as well as the ball's surface.

We can just talk about how the ball moves when it is in the gravitational field of the Earth if we ignore this for the purposes of this discussion. The ball's total energy as it is released is comprised of both its gravitational potential energy and its kinetic energy, which result from the ball's velocity (due to its position).

The gravitational potential energy begins to rise as the ball moves vertically upward at precisely the same pace as it loses kinetic energy. The ball experiences a steady downward acceleration of 9.81 m/s2, which causes it to initially decline until it briefly comes to a stop at its highest point.

Due to its current position in the Earth's gravitational field relative to its initial position, all of the energy at this point is gravitational potential energy. As the ball experiences constant downward acceleration, its motion immediately becomes apparent in that direction because the acceleration easily transforms gravitational potential energy back into kinetic energy.

As a result, at every point along the trajectory, the total of these interchangeable forms of energy remains constant.

To learn more about what happens when a ball is thrown vertically upward:

brainly.com/question/1121850

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A 15 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60° angle with horizontal. (a) Find t
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Answer:

a)    F₁ = 267.3 N,   N₁ = 1300 N,  b)    μ = 0.324

Explanation:

For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces

           

let's use subscript 1 for the ladder and 2 for the firefighter

            ∑ τ = 0

          -W₁ x₁ - W₂ x₂ + N₁ y = 0

           N₁ = \frac{W_1 x_1 + W_2 x_2}{y}          (1)

the center of mass of the ladder is at its geometric center,

d = L / 2 = 15/2 = 7.5 m

         cos 60 = x₁ / d₁

         x₁ = d₁ cos 60

         x₁ = 7.5 cos 60

         x₁ = 3.75 m

for the firefighter d₂ = 4 m

         cos 60 = x₂ / d₂

         x₂ = d₂ cos 60

          x₂ = 4 cos 60 = 2 m

for the fulcrum d₃ = 15 m

         sin 60 = y / d₃

         y = d₃ sin 60

         y = 15 sin 60

         y = 13 m

we look for the Normal by substituting in equation 1

         N₂ = \frac{500 \ 3.75 \ + 800 \ 2}{13}

         N₂ = 267.3 N

now let's use the translational equilibrium relations

 X axis

           F₁ - N₂ = 0

           F₁ = N₂

           F₁ = 267.3 N

Axis y

          N₁ - W₁ -W₂ = 0

          N₁ = W₁ + W₂

          N₁ = 500 + 800

          N₁ = 1300 N

b) for this case change the firefighter's distance d₂ = 9 m

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we substitute in 1

          N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}  

          N₂ = 421.15 N

of the translational equilibrium equation on the x-axis

          fr = F₁ = N₂

          fr = 421.15 N

friction force has the expression

          fr = μ N

in this case the reaction of the Earth to the support of the ladder is N1 = 1300N

          μ = fr / N₁

          μ = 421.15 / 1300

          μ = 0.324

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