Question

Брусок масою 400 г рухається по горизонтальній поверхні під дією сили 1,4 Н. Коефі- цієнт тертя ковзання дорівнює 0,25. 1) З яким прискоренням рухається брусок? 2) У деякій точці швидкість руху бруска дорівнює 4 м/с. Якою буде швидкість руху бруска на відстані 4,5 м від цієї точки (за напрямком руху)?​

Answer 1

Відповідь:

на фото

Пояснення:

можна кращу відповідь;)

Answer 2

Problem Analysis

We are given the following information: - Mass of the block: 400 g - Force acting on the block: 1.4 N - Coefficient of kinetic friction: 0.25

We need to find: 1. The acceleration of the block. 2. The velocity of the block at a distance of 4.5 m from a certain point in the direction of motion.

Solution

1. To find the acceleration of the block, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration: F = ma.

In this case, the net force acting on the block is the force applied minus the force of friction: F_net = F_applied - F_friction.

The force of friction can be calculated using the equation: F_friction = μ * N, where μ is the coefficient of friction and N is the normal force.

Since the block is on a horizontal surface and not accelerating vertically, the normal force is equal to the weight of the block: N = mg, where m is the mass of the block and g is the acceleration due to gravity.

Substituting the values into the equations, we have: - F_net = 1.4 N - (0.25 * 0.4 kg * 9.8 m/s^2) [[1]] - F_net = 1.4 N - 0.98 N - F_net = 0.42 N

Now, we can use Newton's second law to find the acceleration: - F_net = ma - 0.42 N = (0.4 kg) * a - a = 0.42 N / 0.4 kg - a ≈ 1.05 m/s^2

Therefore, the acceleration of the block is approximately 1.05 m/s^2.

2. To find the velocity of the block at a distance of 4.5 m from the given point, we can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.

Since the block starts from rest, the initial velocity u is 0 m/s.

Substituting the values into the equation, we have: - v^2 = 0^2 + 2 * (1.05 m/s^2) * (4.5 m) - v^2 = 0 + 9.45 m^2/s^2 - v ≈ √9.45 m^2/s^2 - v ≈ 3.07 m/s

Therefore, the velocity of the block at a distance of 4.5 m from the given point, in the direction of motion, is approximately 3.07 m/s.

Conclusion

1. The block is moving with an acceleration of approximately 1.05 m/s^2. 2. The velocity of the block at a distance of 4.5 m from the given point, in the direction of motion, is approximately 3.07 m/s.

Please let me know if I can help you with anything else.