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 #3 kg# and speed of #9 m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Answer 1

Julia Aguilar

The compression is #=9m#

The spring constant is #k=1kgs^-2#

The kinetic energy of the object is

#KE=1/2m u^2#

The mass is #m=3kg#

The speed is #u=9ms^-1#

#KE=1/2*3*(9)^2=243/2J#

This kinetic energy will be stored in the spring as potential energy.

#PE=243/2J#

The spring constant is #=3kgs^-2#

So,

#1/2kx^2=243/2#

#x^2=(243)/(3)=81m^2#

#x=sqrt(81)=9m#

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Answer 2

Lily Adams

To find the compression of the spring, use the conservation of mechanical energy equation, $\frac{1}{2}mv^2 = \frac{1}{2}kx^2$ , where $m = 3 \, \text{kg}$ , $v = 9 \, \text{m/s}$ , and $k = 3 \, \text{kg/s}^2$ . Solving for $x$ , you get $x = \sqrt{\frac{mv^2}{k}}$ . Plugging in the values, $x = \sqrt{\frac{(3 \, \text{kg})(9 \, \text{m/s})^2}{3 \, \text{kg/s}^2}} = 9 \, \text{m}$ . Therefore, the spring will compress by $9 \, \text{m}$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.