Question

Шар брошен вертикально вверх со скоростью 25 м/с, на какой высоте кинетическая энергия станет меньше в 3 раза по сравнению с начальной?

Answer 1

Ответ:

20,8 м.

Объяснение:

Кинетическая энергия в начале: $\frac{mV^2}{2}$, потенциальную примем равной нулю. В момент, когда кинетическая энергия станет втрое меньше полная механическая энергия тела будет равна $\frac{mV^2}{6} +mgh$. По закону сохранения энергии:

$\frac{mV^2}{2} =\frac{mV^2}{6} =mgh,$

$\frac{V^2}{3} =gh$,

$h=\frac{V^2}{3g} .$

Вычисляем:

$h=\frac{25^2}{3*10} =20,8(m).$

Answer 2

Calculation of the Height at which the Kinetic Energy becomes 3 Times Smaller

To determine the height at which the kinetic energy becomes three times smaller compared to the initial kinetic energy, we can use the principle of conservation of mechanical energy. The total mechanical energy of an object in free fall consists of its kinetic energy and potential energy.

The formula for kinetic energy is given by:

KE = (1/2)mv^2,

where KE is the kinetic energy, m is the mass of the object, and v is the velocity of the object.

The formula for potential energy is given by:

PE = mgh,

where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height.

According to the principle of conservation of mechanical energy, the sum of the kinetic energy and potential energy remains constant throughout the motion.

Let's denote the initial kinetic energy as KE0 and the final kinetic energy as KEf. We are given that KEf is three times smaller than KE0.

Therefore, we have the equation:

KEf = (1/3)KE0.

Since the kinetic energy is given by (1/2)mv^2, we can rewrite the equation as:

(1/2)mvf^2 = (1/3)(1/2)mv0^2,

where vf is the final velocity and v0 is the initial velocity.

We are given that the initial velocity is 25 m/s. Therefore, we can rewrite the equation as:

(1/2)mvf^2 = (1/3)(1/2)m(25^2).

Simplifying the equation, we have:

vf^2 = (1/3)(25^2).

Taking the square root of both sides, we get:

vf = sqrt((1/3)(25^2)).

Now, we can use the equation for potential energy to find the height at which the kinetic energy becomes three times smaller.

Let's denote the initial height as h0 and the final height as hf. We want to find hf.

We can set up the equation:

KEf + PEf = KE0 + PE0,

where KEf is the final kinetic energy, PEf is the final potential energy, KE0 is the initial kinetic energy, and PE0 is the initial potential energy.

Since we are given that KEf is three times smaller than KE0, we have:

(1/3)KE0 + PEf = KE0 + PE0.

Substituting the formulas for kinetic energy and potential energy, we have:

(1/3)(1/2)mvf^2 + mghf = (1/2)mv0^2 + mgh0.

Since we know the values of vf, v0, and h0, we can solve for hf.

Let's substitute the values and calculate the height.

m = 1 (assuming the mass of the object is 1 kg) vf = sqrt((1/3)(25^2)) v0 = 25 m/s h0 = ?

Substituting the values into the equation, we have:

(1/3)(1/2)(1)(sqrt((1/3)(25^2)))^2 + ghf = (1/2)(1)(25^2) + g(0).

Simplifying the equation, we get:

(1/3)(1/2)(1)(1/3)(25^2) + ghf = (1/2)(1)(25^2).

Simplifying further, we have:

(1/18)(25^2) + ghf = (1/2)(25^2).

Simplifying again, we get:

(1/18)(625) + ghf = (1/2)(625).

Now, we can solve for hf.

Let's calculate the height.