
Под действием постоянной силы 5 н тело начинает двигаться с ускорением 0,2 м/с. какую работу
совершит эта сила за первые 20 секунд действия

Ответы на вопрос

A = 5 ·



The user's message is a physics question in Russian. It asks: "Answer in detail. Under the action of a constant force of 5 N, a body begins to move with an acceleration of 0.2 m/s. How much work will this force do in the first 20 seconds of action?"
To answer this question, we need to use the formula for the work done by a constant force:
$$W = F \cdot s$$
where $W$ is the work, $F$ is the force, and $s$ is the displacement of the body.
To find the displacement of the body, we need to use the formula for the motion of a body with constant acceleration:
$$s = v_0 \cdot t + \frac{a \cdot t^2}{2}$$
where $s$ is the displacement, $v_0$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.
Since the body begins to move under the action of the force, we can assume that its initial velocity is zero. Therefore, the formula for the displacement simplifies to:
$$s = \frac{a \cdot t^2}{2}$$
Substituting the given values of $a$ and $t$, we get:
$$s = \frac{0.2 \cdot 20^2}{2}$$
$$s = 40 \text{ m}$$
Now, we can plug this value of $s$ into the formula for the work done by the force:
$$W = F \cdot s$$
$$W = 5 \cdot 40$$
$$W = 200 \text{ J}$$
Therefore, the work done by the force in the first 20 seconds of action is 200 J.
I hope this answer helps you understand the problem. If you want to learn more about the concepts of force, work, and energy, you can check out some of the web search results that I found for you:
- [A tutorial on force and work](https://reshak.ru/otvet/otvet12.php?otvet1=322) - [A video lesson on work and energy](https://uchitel.pro/%d0%b7%d0%b0%d0%b4%d0%b0%d1%87%d0%b8-%d0%bd%d0%b0-%d0%b7%d0%b0%d0%ba%d0%be%d0%bd%d1%8b-%d0%bd%d1%8c%d1%8e%d1%82%d0%be%d0%bd%d0%b0/) - [An interactive simulation on work and energy](https://uchi.ru/otvety/questions/pod-deystviem-postoyannoy-sili-5-n-telo-nachinaet-dvigatsya-s-uskoreniem-02-m-s-kakuyu-ra)


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