Question

На столе лежит гиря массой 2кг с каким ускорением может двигаться эта гиря если подействовать на нее в горизонтальном напроалении силой 20 ньютон ? кофицент трения гири о поверхность стола равен 0.5

Answer 1

F=ma Fтр=n*m*g=.5*2*10=10 F-Fтр=ma a=(F-Fтр)/m=(20-10)/2=5м/с^2

Answer 2

Calculation of Acceleration

To calculate the acceleration of the 2 kg weight when a force of 20 Newton is applied horizontally and the coefficient of friction between the weight and the table surface is 0.5, we can use Newton's second law of motion.

Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, it can be expressed as:

F = m * a

Where: - F is the net force acting on the object, - m is the mass of the object, and - a is the acceleration of the object.

In this case, the net force acting on the weight is the applied force minus the force of friction. The force of friction can be calculated by multiplying the coefficient of friction by the normal force, which is equal to the weight of the object (mass * acceleration due to gravity).

Let's calculate the acceleration step by step:

1. Calculate the force of friction: - Force of friction = coefficient of friction * normal force - Normal force = mass * acceleration due to gravity

Given: - Mass of the weight (m) = 2 kg - Coefficient of friction (μ) = 0.5 - Acceleration due to gravity (g) = 9.8 m/s^2

Calculations: - Normal force = 2 kg * 9.8 m/s^2 = 19.6 N - Force of friction = 0.5 * 19.6 N = 9.8 N

2. Calculate the net force: - Net force = applied force - force of friction

Given: - Applied force = 20 N

Calculation: - Net force = 20 N - 9.8 N = 10.2 N

3. Calculate the acceleration: - Acceleration = net force / mass

Given: - Mass of the weight (m) = 2 kg - Net force = 10.2 N

Calculation: - Acceleration = 10.2 N / 2 kg = 5.1 m/s^2

Therefore, the 2 kg weight can move with an acceleration of 5.1 m/s^2 when a force of 20 N is applied horizontally and the coefficient of friction between the weight and the table surface is 0.5.

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