Question

Автомобиль массой 2т поднимается в гору с уклоном 0,2. На участке пути 32м скорость автомобиля возросла от 21,6км/ч до 36км/ч. Считая движение автомобиля равноускоренным, определить силу тяги двигателя. Коэффициент трения равен 0,02.

Answer 1

Задачу можно решать с использованием закона сохранения энергии: A=0,5*(m*v^2-m*v0^2)+m*g*h+Ft*S; h=S*sin(a); Ft=k*m*g*cos(a); A=F*S;

F*S=0,5*(m*v^2-m*v0^2)+m*g*S*sin(a)+k*m*g*cos(a)*S; m=2000; v=6; v2=10; g=9,8;

sin(a)=0,2; cos(a)=1(примерно);k=0,02.Считайте!

Answer 2

I found some web search results that might help you with your question. Here is a detailed answer based on one of the sources:

- First, we need to convert the speeds from km/h to m/s by dividing them by 3.6. So, v1 = 21.6 / 3.6 = 6 m/s and v2 = 36 / 3.6 = 10 m/s. - Next, we need to find the acceleration of the car by using the formula S = (v2^2 - v1^2) / (2 * a), where S is the distance, v1 and v2 are the initial and final speeds, and a is the acceleration. Rearranging the formula, we get a = (v2^2 - v1^2) / (2 * S) = (10^2 - 6^2) / (2 * 32) = 1 m/s^2. - Then, we need to find the net force acting on the car by using the formula ma = F - mg * sin(alpha) - mu * mg * cos(alpha), where m is the mass, a is the acceleration, F is the force of the engine, g is the gravitational acceleration, alpha is the slope angle, and mu is the coefficient of friction. We can find the slope angle by using the formula tan(alpha) = 0.2, which gives alpha = 11.31 degrees. Substituting the values, we get 2000 * 1 = F - 2000 * 10 * sin(11.31) - 0.02 * 2000 * 10 * cos(11.31). - Finally, we need to solve for F by rearranging the equation. We get F = 2000 * 1 + 2000 * 10 * sin(11.31) + 0.02 * 2000 * 10 * cos(11.31) = 6391.92 N. This is the force of the engine that pulls the car up the hill.

I hope this answer helps you understand the problem better. If you want to see more sources, you can check out the other web search results .