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 #3 kg# and speed of #1 m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Question

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 #3  kg# and speed of #1  m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Isabella Alvarez · Accepted Answer

#=0.87m# By conservation of mechanical energy when the spring is compressed fully after collision PE gained by the spring = Initial KE of the colliding object #=>cancel (1/2)kx^2=cancel(1/2)mv^2# where m = mass=3kg v=velocity of the object=#1m/s# k = force constant=#4(kg)/s^2# x= compression of spring=? #=>x^2=mv^2/k# #=>x=sqrt(m/kv^2)=sqrt(m/k)xxv=sqrt((3kg)/(4kg)/s^2)xx1m/s=0.87m#

Charlotte Aaron · Answer

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

The initial kinetic energy of the object is $ KE = \frac{1}{2}mv^2 $, where $ m = 3 $ kg and $ v = 1 $ m/s.

Substituting the values, we get:
$$ KE = \frac{1}{2}(3 \, \text{kg})(1 \, \text{m/s})^2 = \frac{1}{2}(3)(1) = 1.5 \, \text{J} $$

Since the object comes to rest, all of its initial kinetic energy is converted into the potential energy stored in the spring. The potential energy stored in the spring is given by $ PE = \frac{1}{2}kx^2 $, where $ k = 4 \, \text{kg/s}^2 $ (the spring constant) and $ x $ is the compression of the spring.

Setting the initial kinetic energy equal to the potential energy stored in the spring:
$$ 1.5 \, \text{J} = \frac{1}{2}(4 \, \text{kg/s}^2)x^2 $$

Solving for $ x $:
$$ x^2 = \frac{1.5 \, \text{J}}{\frac{1}{2}(4 \, \text{kg/s}^2)} = \frac{1.5 \, \text{J}}{2 \, \text{N/m}} = 0.75 \, \text{m}^2 $$

$$ x = \sqrt{0.75 \, \text{m}^2} \approx 0.866 \, \text{m} $$

So, the spring will compress approximately 0.866 meters.

Amelia Adams · Answer

undefined The spring compression can be calculated using the conservation of mechanical energy principle. The initial kinetic energy of the object is converted into potential energy stored in the spring when it comes to a stop. The equation to find the compression of the spring ($x$) is given by:

$$ \frac{1}{2}mv^2 = \frac{1}{2}kx^2 $$

where:
$ m $ = mass of the object (3 kg)
$ v $ = initial velocity of the object (1 m/s)
$ k $ = spring constant (4 kg/s$^2$)
$ x $ = compression of the spring

Plugging in the given values:

$$ \frac{1}{2}(3)(1)^2 = \frac{1}{2}(4)x^2 $$

Solving for $ x $:

$$ x^2 = \frac{3}{4} $$

$$ x = \sqrt{\frac{3}{4}} $$

$$ x = \frac{\sqrt{3}}{2} $$

Therefore, the spring compresses by $ \frac{\sqrt{3}}{2} $ meters.