A spring with a constant of #3 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #5 kg# and speed of #8 m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

To find the compression of the spring, we can use the principle of conservation of mechanical energy. The initial kinetic energy of the object is converted into potential energy stored in the spring when it compresses.

The formula for the potential energy stored in a spring is:

(PE = \frac{1}{2}kx^2)

Where: (PE) = potential energy stored in the spring (in joules) (k) = spring constant (in (kg/s^2)) (x) = compression distance (in meters)

Given: (k = 3 , \text{kg/s}^2) (m = 5 , \text{kg}) (v = 8 , \text{m/s})

Using the formula for kinetic energy:

(KE = \frac{1}{2}mv^2)

The initial kinetic energy of the object before collision is:

(KE_{initial} = \frac{1}{2} \times 5 , \text{kg} \times (8 , \text{m/s})^2) (KE_{initial} = 160 , \text{J})

Since the object stops moving after compressing the spring, all of its initial kinetic energy is converted into potential energy stored in the spring. Thus, we can equate the two:

(PE = KE_{initial})

Substituting the values:

(\frac{1}{2} \times 3 , \text{kg/s}^2 \times x^2 = 160 , \text{J})

Solving for (x):

(x^2 = \frac{160 , \text{J}}{\frac{1}{2} \times 3 , \text{kg/s}^2}) (x^2 = \frac{320 , \text{kg} \cdot \text{m}^2}{\text{kg/s}^2}) (x^2 = \frac{320 , \text{N} \cdot \text{m}}{\text{N}}) (x^2 = 320 , \text{m}) (x \approx 17.89 , \text{m})

So, the spring will compress approximately (17.89 , \text{m}).