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meriva
3 years ago
15

In a physics lab experiment, a compressed spring launches a 24 g metal ball at a 35o angle above the horizontal. Compressing the

spring 17cm causes the ball to hit the floor 2.0 m below the point at which it leaves the spring after traveling 5.3 m horizontally. What is the spring constant
Physics
1 answer:
Levart [38]3 years ago
4 0

Answer:

k = 45.95 N/m

Explanation:

First, we will find the launch speed of the ball by using the formula for the horizontal range of the projectile.

R = \frac{v_{o}^{2}\ Sin\ 2\theta}{g} \\\\v_{o}^{2} = \frac{Rg}{Sin\ 2\theta}\\

where,

Vo = Launch Speed = ?

R = Horizontal Range = 5.3 m

θ = Launch Angle = 35°

Therefore,

v_{o}^{2} = \frac{(5.3\ m)(9.81\ m/s^{2})}{Sin\ 2(35^{o})}\\

v₀² = 55.33 m²/s²

Now, we know that the kinetic energy gain of ball is equal to the potential energy stored by spring:

Kinetic\ Energy\ Gained\ By\ Ball = Elastic\ Potential\ Energy\ Stored\ in \ Spring\\\frac{1}{2}mv_{o}^{2} = \frac{1}{2}kx^{2}\\\\k = \frac{mv_{o}^{2}}{x^2} \\

where,

k = spring constant = ?

x = compression = 17 cm = 0.17 m

m = mass of ball = 24 g = 0.024 kg

Therefore,

k = \frac{(0.024\ kg)(55.33\ m^2/s^2)}{(0.17\ m)^2} \\

<u>k = 45.95 N/m</u>

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