Question

Брусок масою 5 кг притиснуто до стіни. Під дією зовнішньої сили , що прикладено як зображено на рисунку, рухається з постійним прискоренням 1 м\с2. Визначте величину прикладеної сили , якщо коефіцієнт тертя між стінкою та бруском 0,2

Answer 1

To determine the magnitude of the applied force, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the mass of the block is 5 kg and the acceleration is 1 m/s^2.

Using the formula F = m * a, we can calculate the force:

F = 5 kg * 1 m/s^2 F = 5 N

Therefore, the magnitude of the applied force is 5 Newtons.

Now, let's consider the coefficient of friction between the wall and the block, which is given as 0.2. The frictional force can be calculated using the formula F_friction = coefficient of friction * normal force.

The normal force is equal to the weight of the block, which can be calculated using the formula F_weight = mass * gravitational acceleration.

Assuming the gravitational acceleration is approximately 9.8 m/s^2, we can calculate the weight of the block:

F_weight = 5 kg * 9.8 m/s^2 F_weight = 49 N

Now, we can calculate the frictional force:

F_friction = 0.2 * 49 N F_friction = 9.8 N

Therefore, the magnitude of the frictional force is 9.8 Newtons.

Please note that the direction of the applied force and the frictional force may vary depending on the specific scenario and the orientation of the block.