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 #7 kg # and speed of # 8 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 #7   kg # and speed of # 8  m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Charles Aeschliman · Accepted Answer

The spring will compress by #=10.58m#

The spring constant is #k=4kgs^-2#

The kinetic energy of the object is

#KE=1/2m u^2#

#KE=1/2*7*8^2=224J#

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

#PE=224J#

So,

#1/2kx^2=224#

#x^2=448/k=448/4=112#

#x=sqrt112=10.58m#

Adrian Ayala · 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 compressed spring.

Initial kinetic energy of the object = 1/2 * mass * velocity^2
= 1/2 * 7 kg * (8 m/s)^2
= 224 J

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

Equating the initial kinetic energy to the potential energy stored in the compressed spring:

224 J = 1/2 * 4 (kg/s^2) * compression^2

Solving for compression:

compression^2 = (224 J * 2) / 4 (kg/s^2)
compression^2 = 448 J / 4 (kg/s^2)
compression^2 = 112 m^2

compression = sqrt(112 m^2)
compression ≈ 10.58 m

So, the spring will compress by approximately 10.58 meters.