Question

Брусок массой 2 кг скользит вниз по наклонной плоскости с ускорением 2 м/с2.угол наклона плоскости 30 градусов.изобразите на чертеже все силы,действующие на брусок ,и определите чему равна равнодействующая этих сил. Ребят,очень нужна помощь в решении задач по физике,кто хорошо понимает физику.отпишитесь,пожалуйста!!!

Answer 1

Коэффициент трения не дан, значит его учитывать не будем
запишем 2 закон Ньютона
в векторном виде
$ma=N+mg$
$ma$  это и есть та самая результирующая сила
тк нам все дано $ma=2*4=4N$ 

силы на рисунке
сила тяжести $mg$ направлена вниз к центру Земли
реакции опоры $N$ направлена вверх перпендикулярно поверхности

Answer 2

Problem Analysis

We have a block with a mass of 2 kg sliding down a inclined plane with an acceleration of 2 m/s^2. The angle of inclination of the plane is 30 degrees. We need to draw a diagram showing all the forces acting on the block and determine the resultant force.

Diagram

To better understand the problem, let's draw a diagram representing the situation:

          |\
          | \
          |  \
          |   \
          |    \
          |     \
          |      \
          |       \
          |        \
          |         \
          |          \
          |           \
          |            \
          |             \
          |              \
          |               \
          |                \
          |                 \
          |                  \
          |___________________\

In the diagram, the inclined plane is represented by the diagonal line, and the block is sliding down the plane.

Forces Acting on the Block

There are several forces acting on the block as it slides down the inclined plane. Let's identify and label them:

1. Weight (mg): This is the force due to gravity acting vertically downward. Its magnitude can be calculated using the formula: mg, where m is the mass of the block and g is the acceleration due to gravity. In this case, the mass of the block is 2 kg, so the weight is 2 * 9.8 = 19.6 N. 2. Normal Force (N): This is the force exerted by the inclined plane on the block perpendicular to the plane. It counteracts the component of the weight that is perpendicular to the plane. Its magnitude is equal to the component of the weight perpendicular to the plane, which can be calculated using the formula: N = mg * cos(theta), where theta is the angle of inclination of the plane. In this case, the angle of inclination is 30 degrees, so the normal force is 19.6 * cos(30) = 16.96 N. 3. Friction Force (F): This is the force opposing the motion of the block along the inclined plane. Its magnitude can be calculated using the formula: F = mu * N, where mu is the coefficient of friction between the block and the plane. However, since the block is sliding down the plane, the friction force is directed opposite to the motion and can be calculated as F = mu * N * -1. The coefficient of friction is not given in the problem statement, so we cannot determine its exact value. 4. Resultant Force (R): This is the net force acting on the block in the direction of motion. It is the vector sum of all the forces acting on the block. Its magnitude can be calculated using the formula: R = ma, where m is the mass of the block and a is the acceleration. In this case, the mass of the block is 2 kg and the acceleration is 2 m/s^2, so the resultant force is 2 * 2 = 4 N.

Conclusion

In conclusion, the forces acting on the block sliding down the inclined plane are the weight (mg), the normal force (N), and the friction force (F). The resultant force (R) is equal to 4 N.

Please note that the coefficient of friction was not provided in the problem statement, so we cannot determine its exact value.