Question

По столу движется брусок массой 250 г под действием силы, равной по модулю 1 Н и приложенной вверх под углом 30 градусов к горизонту. Коэффициент трения между бруском и столом 0,3. Чему равно ускорение бруска?

Answer 1

На тело действуют следующие силы:
- сила тяжести mg - вертикально вниз
- сила реакции опоры N - вертикально вверх
- сила трения Fтр=μN - назад, параллельно поверхности
- сила, приложенная к телу под углом к горизонту - F.
Запишем уравнение движения в проекциях на горизонтальную и вертикальную ось.
1) Горизонтальная:
ma=Fcos(a)-Fтр=Fcos(a)-μN
2) Вертикальная: 
N+Fsin(a)=mg, откуда N=mg-Fsin(a)
Подставляя второе равенство в первое, получаем:
ma=Fcos(a)-μ(mg-Fsin(a))
a=F(cos(a)+μsin(a))/m-μg
a=1*(√3/2+0,3*0,5)/0,25-0,3*10≈1 м/с^2

Answer 2

Calculation of the Acceleration of the Block

To calculate the acceleration of the block, we need to consider the forces acting on it. The forces involved are the applied force, the force of gravity, and the force of friction.

1. Force of gravity (weight): The force of gravity acting on the block can be calculated using the formula F = 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 given as 250 g, which is equivalent to 0.25 kg. The acceleration due to gravity is approximately 9.8 m/s^2.

Therefore, the force of gravity acting on the block is F = (0.25 kg) * (9.8 m/s^2) = 2.45 N. 2. Applied force: The applied force is given as 1 N and is directed upward at an angle of 30 degrees to the horizontal.

To find the horizontal component of the applied force, we can use the formula Fx = F * cos(theta), where F is the magnitude of the force and theta is the angle between the force and the horizontal direction. In this case, F = 1 N and theta = 30 degrees.

Therefore, the horizontal component of the applied force is Fx = (1 N) * cos(30 degrees) = 0.87 N.

To find the vertical component of the applied force, we can use the formula Fy = F * sin(theta), where F is the magnitude of the force and theta is the angle between the force and the horizontal direction. In this case, F = 1 N and theta = 30 degrees.

Therefore, the vertical component of the applied force is Fy = (1 N) * sin(30 degrees) = 0.5 N.

3. Force of friction: The force of friction can be calculated using the formula Ff = μ * N, where μ is the coefficient of friction and N is the normal force. The normal force is equal to the force of gravity acting on the block.

In this case, the coefficient of friction between the block and the table is given as 0.3.

Therefore, the force of friction is Ff = (0.3) * (2.45 N) = 0.735 N.

4. Net force: The net force acting on the block is the vector sum of the applied force and the force of friction. Since the applied force is directed upward and the force of friction is directed opposite to the motion, the net force can be calculated as:

Fnet = Fy - Ff = 0.5 N - 0.735 N = -0.235 N.

The negative sign indicates that the net force is acting in the opposite direction to the motion.

5. Acceleration: The acceleration of the block can be calculated using Newton's second law, which states that the net 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 0.25 kg.

Therefore, the acceleration of the block is:

a = Fnet / m = (-0.235 N) / (0.25 kg) = -0.94 m/s^2.

The negative sign indicates that the block is accelerating in the opposite direction to the applied force.

Answer:

The acceleration of the block is approximately -0.94 m/s^2.