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 #6 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 #3 (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 #8  m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Elena Albarran · Accepted Answer

#11.3cm(1dp)#

We need the following formulae

Elastic potential energy for springs

#EPE=1/2kx^2#

where#k= "the spring constant; "#

#x="the compression/extension of the spring"#

Kinetic energy

#KE=1/2mv^2#

#m="the mass of the object;" #

# v="the speed of the object"#

we have an object of mass#" "6kg" "#moving with speed #8ms^(-1)#

it will have kinetic energy of

#KE=1/2mv^2=1/2xx6xx8^2#

#KE=192J#

on collision with the spring all this will be given to the spring in terms of EPE

#KE=EPE#

#192=1/2xx3xxx^2#

#x^2=(2xx192)/3=128#

#:.x=sqrt(128)=11.3cm(1dp)#

Elena Albarran · Answer

undefined To find the compression of the spring, you can use the principle of conservation of mechanical energy. The initial kinetic energy of the object will be equal to the potential energy stored in the compressed spring.

The formula for potential energy stored in a compressed spring is:

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

Where:
- $ U $ is the potential energy stored in the spring,
- $ k $ is the spring constant (3 kg/s² in this case),
- $ x $ is the compression of the 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 (6 kg),
- $ v $ is the velocity of the object (8 m/s).

Set the initial kinetic energy equal to the potential energy of the spring and solve for $ x $.

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

$$ 6 \times 8^2 = 3 \times x^2 $$

$$ 6 \times 64 = 3x^2 $$

$$ 384 = 3x^2 $$

$$ x^2 = \frac{384}{3} $$

$$ x^2 = 128 $$

$$ x = \sqrt{128} $$

$$ x \approx 11.31 \text{ meters} $$

So, the spring will compress approximately 11.31 meters.