# A spring with a constant of #4 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #6 kg# and speed of #3 m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

3.7m

So the spring will move 3.7m.

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To find the compression of the spring, you can use the principle of conservation of mechanical energy. The initial kinetic energy of the object is equal to the potential energy stored in the compressed spring. The initial kinetic energy of the object is given by ( KE = \frac{1}{2}mv^2 ), where ( m ) is the mass of the object and ( v ) is its speed. The potential energy stored in the compressed spring is given by ( PE = \frac{1}{2}kx^2 ), where ( k ) is the spring constant and ( x ) is the compression of the spring. Setting these two equal and solving for ( x ), you get:

[ \frac{1}{2}mv^2 = \frac{1}{2}kx^2 ] [ x = \sqrt{\frac{mv^2}{k}} ]

Substituting the given values, ( m = 6 , \text{kg} ), ( v = 3 , \text{m/s} ), and ( k = 4 , \text{kg/s}^2 ), you get:

[ x = \sqrt{\frac{(6 , \text{kg})(3 , \text{m/s})^2}{4 , \text{kg/s}^2}} ] [ x = \sqrt{\frac{54 , \text{kg m}^2/\text{s}^2}{4 , \text{kg/s}^2}} ] [ x = \sqrt{\frac{54 , \text{m}^2}{4}} ] [ x = \sqrt{13.5} , \text{m} ] [ x \approx 3.674 , \text{m} ]

So, the spring will compress approximately ( 3.674 , \text{m} ).

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