# 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?

Aug 25, 2017

$11.3 c m \left(1 \mathrm{dp}\right)$

#### Explanation:

We need the following formulae

Elastic potential energy for springs

$E P E = \frac{1}{2} k {x}^{2}$

where$k = \text{the spring constant; }$

$x = \text{the compression/extension of the spring}$

Kinetic energy

$K E = \frac{1}{2} m {v}^{2}$

$m = \text{the mass of the object;}$

$v = \text{the speed of the object}$

we have an object of mass$\text{ "6kg" }$moving with speed $8 m {s}^{- 1}$

it will have kinetic energy of

$K E = \frac{1}{2} m {v}^{2} = \frac{1}{2} \times 6 \times {8}^{2}$

$K E = 192 J$

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

$K E = E P E$

$192 = \frac{1}{2} \times 3 \times {x}^{2}$

${x}^{2} = \frac{2 \times 192}{3} = 128$

$\therefore x = \sqrt{128} = 11.3 c m \left(1 \mathrm{dp}\right)$