Question

Сжатая на 5см легкая пружина, один конец которой прикреплен к стене, распрямляется и толкает шарик массой 20г в горизонтальном направлении. Какую скорость приобретет шарик, если жесткость пружины равна 800н/м?

Answer 1

kx^2/2=mV^2/2

$V=\sqrt{kx^{2}/2$$V=\sqrt{800*0,05*0.05 / 0,02}=10$ м|c

Answer 2

Calculation of the Ball's Velocity

To calculate the velocity acquired by the ball when pushed by the compressed spring, we can use the principle of conservation of mechanical energy. The potential energy stored in the compressed spring is converted into kinetic energy as the ball moves.

The potential energy stored in the spring can be calculated using the formula:

Potential Energy (PE) = (1/2) * k * x^2

Where: - k is the spring constant (stiffness) in N/m, which is given as 800 N/m in this case. - x is the compression or extension of the spring in meters, which is given as 0.05 m (5 cm) in this case.

Substituting the given values into the formula, we can calculate the potential energy:

PE = (1/2) * 800 N/m * (0.05 m)^2

Simplifying the equation:

PE = 0.5 * 800 N/m * 0.0025 m^2

PE = 0.5 * 800 * 0.0025 J

PE = 1 J

The potential energy stored in the spring is 1 Joule.

According to the principle of conservation of mechanical energy, this potential energy is converted into kinetic energy as the ball moves. The kinetic energy of an object can be calculated using the formula:

Kinetic Energy (KE) = (1/2) * m * v^2

Where: - m is the mass of the ball in kilograms, which is given as 20 g (0.02 kg) in this case. - v is the velocity of the ball in m/s, which we need to calculate.

Since the potential energy is converted into kinetic energy, we can equate the two equations:

PE = KE

1 J = (1/2) * 0.02 kg * v^2

Simplifying the equation:

1 J = 0.01 kg * v^2

Rearranging the equation to solve for v:

v^2 = (1 J) / (0.01 kg)

v^2 = 100 m^2/s^2

Taking the square root of both sides:

v = 10 m/s

Therefore, the ball will acquire a velocity of 10 m/s when pushed by the compressed spring with a stiffness of 800 N/m.