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

Leo Abeyta · Accepted Answer

#Delta x=1m# #"the Kinetic Energy of an object that has a mass of m and velocity of v is given by:"# #E_k=1/2.m.v^2# #E_k=1/2*3*1^2=1/2 *3J# #E_p=E_k# #"it will gain a potential energy if spring compresses"# #E_p = 1/2.k. Delta x^2# #cancel(1/2) 3* Delta x^2=cancel(1/2) *3# #Delta x^2=1# #Delta x=1m#

Leo Abeyta · Answer

undefined To find the compression of the spring, we can use the conservation of mechanical energy. The initial kinetic energy of the object is equal to the potential energy stored in the spring when it stops moving.

Initial kinetic energy = 1/2 * mass * velocity^2
Initial kinetic energy = 1/2 * 3 kg * (1 m/s)^2
Initial kinetic energy = 1.5 J

Potential energy stored in the spring = 1/2 * spring constant * compression^2

1.5 J = 1/2 * 3 (kg/s^2) * compression^2

Solving for compression:

compression^2 = (1.5 J * 2) / (3 kg/s^2)
compression^2 = 1 J / (3 kg/s^2)
compression^2 = 1/3 m^2

compression = sqrt(1/3) m
compression ≈ 0.577 m

So, the spring will compress approximately 0.577 meters.

Isabella Alvarez · Answer

undefined To determine how much the spring will compress, you can use the principle of conservation of mechanical energy. When the object collides with the spring, its initial kinetic energy will be converted into potential energy stored in the compressed spring.

The equation for the potential energy stored in a compressed spring is given by:

$$ PE = \frac{1}{2} k x^2 $$

Where:
- $ PE $ is the potential energy stored in the spring,
- $ k $ is the spring constant (given as $ 3 \, \text{kg/s}^2 $),
- $ x $ is the displacement or compression of the spring from its equilibrium position.

The kinetic energy of the object just before it collides with the spring is given by:

$$ KE = \frac{1}{2} m v^2 $$

Where:
- $ KE $ is the kinetic energy of the object,
- $ m $ is the mass of the object (given as $ 3 \, \text{kg} $),
- $ v $ is the speed of the object (given as $ 1 \, \text{m/s} $).

Since mechanical energy is conserved, the initial kinetic energy of the object will be equal to the potential energy stored in the compressed spring:

$$ \frac{1}{2} m v^2 = \frac{1}{2} k x^2 $$

Rearranging the equation to solve for $ x $:

$$ x = \sqrt{\frac{m v^2}{k}} $$

Substituting the given values:

$$ x = \sqrt{\frac{(3 \, \text{kg})(1 \, \text{m/s})^2}{3 \, \text{kg/s}^2}} $$

$$ x = \sqrt{\frac{3 \, \text{kg} \cdot \text{m}^2/\text{s}^2}{3 \, \text{kg/s}^2}} $$

$$ x = \sqrt{1 \, \text{m}^2} $$

$$ x = 1 \, \text{m} $$

Therefore, the spring will compress by $ 1 \, \text{meter} $.