Правка разделов «Solution», «Answer: 1. Q = m(0.5V²-gh) 2. Q = m(0.5v»
en/2.4.33.md
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| − | ### Statement | ||
| + | $2.4.33.$ __"Amount of heat"__ | ||
| − | $2.4.33.$ [Insert problem description here] | ||
| + | __Statement__: | ||
| + | $2.4.33.$ A body of mass $m$ was pushed up an inclined plane. After wich it moved with an initial speed and then stopped, rising the height $h$. What amount of heat was released in this case. | ||
| − | __Example Statement__: | ||
| − | $1.1.1.$ Determine the coordinate $x(t)$ of a body as a function of time $t$, given that its acceleration is defined as $a(t) = bt$, where $b$ is a constant. | ||
| − | |||
| ### Solution | |||
| − | [Your solution should be placed here] | ||
| + |  | ||
| − | __Example Solution__: | ||
| − | The acceleration of the body defined by | ||
| − | $$a(t) = bt$$ | ||
| − | We know that acceleration is the time derivative of velocity: | ||
| − | $$a(t) = \frac{d v(t)}{d t}$$ | ||
| − | |||
| − | To find the velocity $v(t)$, we integrate $a(t)$ with respect to time: | ||
| − | |||
| − | $$v(t) = \int a(t) \, dt = \int b t \, dt$$ | ||
| − | |||
| − | If the initial velocity is $v(0) = 0$, then the velocity becomes: | ||
| − | |||
| − | $$v(t) = \frac{b t^2}{2}$$ | ||
| − | |||
| − | Likewise, integrate $v(t)$ with respect to time: | ||
| − | |||
| − | $$x(t)= \int v(t) \, dt = \frac{b}{2} \int t^2 \, dt$$ | ||
| − | |||
| − | From where the coordinate from time, considering the initial conditions: | ||
| − | |||
| − | $$\boxed{x(t)=\frac{bt^3}{6}}$$ | ||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result, like this:] | ||
| − | |||
| − | |||
| − | __Example Answer__: | ||
| − | $$ x(t)=\frac{bt^3}{6} $$ | ||
| + | ### Answer: 1. Q = m(0.5V²-gh) 2. Q = m(0.5v²-gh(1+μcotgα)) | ||
| @@ -1,44 +1,14 @@ | |||
| ### Statement | $2.4.33.$ __"Amount of heat"__ | ||
| $2.4.33.$ [Insert problem description here] | __Statement__: | ||
| $2.4.33.$ A body of mass $m$ was pushed up an inclined plane. After wich it moved with an initial speed and then stopped, rising the height $h$. What amount of heat was released in this case. | |||
| __Example Statement__: | |||
| $1.1.1.$ Determine the coordinate $x(t)$ of a body as a function of time $t$, given that its acceleration is defined as $a(t) = bt$, where $b$ is a constant. | |||
| ### Solution | ### Solution | ||
| [Your solution should be placed here] |  | ||
| __Example Solution__: | |||
| The acceleration of the body defined by | |||
| $$a(t) = bt$$ | |||
| We know that acceleration is the time derivative of velocity: | |||
| $$a(t) = \frac{d v(t)}{d t}$$ | ### Answer: 1. Q = m(0.5V²-gh) 2. Q = m(0.5v²-gh(1+μcotgα)) | ||
| To find the velocity $v(t)$, we integrate $a(t)$ with respect to time: | |||
| $$v(t) = \int a(t) \, dt = \int b t \, dt$$ | |||
| If the initial velocity is $v(0) = 0$, then the velocity becomes: | |||
| $$v(t) = \frac{b t^2}{2}$$ | |||
| Likewise, integrate $v(t)$ with respect to time: | |||
| $$x(t)= \int v(t) \, dt = \frac{b}{2} \int t^2 \, dt$$ | |||
| From where the coordinate from time, considering the initial conditions: | |||
| $$\boxed{x(t)=\frac{bt^3}{6}}$$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result, like this:] | |||
| __Example Answer__: | |||
| $$ x(t)=\frac{bt^3}{6} $$ | |||