In a physics lab experiment, a spring clamped to the table shoots a 22 g ball horizontally. When the spring is compressed 22 cm, the ball travels horizontally 5.2 m and lands on the floor 1.3 m below the point at which it left the spring. What is the spring constant?

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??? someone please answer with a step by step how to, i have mastering physics too :(

To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position.

Let's break down the given information:

Mass of the ball (m): 22 g = 0.022 kg
Compression distance of the spring (x): 22 cm = 0.22 m
Horizontal distance traveled by the ball (d): 5.2 m
Vertical distance fallen by the ball (h): 1.3 m

In this scenario, we can assume that the horizontal velocity of the ball remains constant throughout its motion.

First, we need to calculate the launch velocity (v) of the ball using the horizontal distance traveled:

v = d / t

Since the horizontal distance traveled by the ball is given and it was shot horizontally, we can assume there is no horizontal acceleration. Thus, the time taken (t) to travel the horizontal distance is the same as it would be for a freely falling object.

To calculate the time of flight (t), we can use the free fall equation:

h = (1/2) * g * t^2

where g is the acceleration due to gravity (approximately 9.8 m/s^2). Rearranging the equation and solving for t:

t = sqrt((2 * h) / g)

Substituting the given values:

t = sqrt((2 * 1.3) / 9.8) ≈ 0.51 s

Now, we can calculate the launch velocity (v):

v = d / t
v = 5.2 / 0.51 ≈ 10.2 m/s

Since the horizontal velocity of the ball is constant, it is also equal to the velocity of the mass on the spring (v_m). This velocity can be related to the compression distance (x) and the spring constant (k) using the equation:

v_m = sqrt((k / m) * x^2)

Rearranging the equation and solving for the spring constant (k):

k = (m / x^2) * v_m^2

Substituting the given values:

k = (0.022 / 0.22^2) * 10.2^2 ≈ 261 N/m

Therefore, the spring constant of the clamped spring in this experiment is approximately 261 N/m.