Question

A 6.0-kg box moving at 3.0 m/s on a horizontal, friction-less surface runs into a light spring of force constant 75 N/cm. Use the work—energy theorem to find the maximum compression of the spring.

Answer 1

Answer:

The maximum compression in the spring is 0.085 m or 8.5 cm.

Explanation:

Given:

Mass of box (m) = 6.0 kg

Spring constant of spring (k) = 75 N/cm = 75 × 100 N/m = 7500 N/m

Initial velocity of the box (u) = 3.0 m/s

Final velocity at maximum compression (v) = 0 m/s

Let the maximum compression be 'x'.

Surface is frictionless. So, only spring force is the force acting on the box spring system.

Now, as per work-energy theorem, the work done by the net force acting on a body is equal to the change in the kinetic energy of the body.

Here, the work done by the spring force is given as:

$W_{net}=-\frac{1}{2}kx^2$

The negative sign implies that force and displacement are in opposite direction.

The change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)$

Now, from work-energy theorem:

$W_{net}=\Delta K\\\\-\frac{1}{2}kx^2=\frac{1}{2}m(v^2-u^2)\\\\x^2=\frac{m(v^2-u^2)}{-k}\\\\x=\sqrt{\frac{m(v^2-u^2)}{-k}}$

Now, plug in the given values and solve for 'x'. This gives,

$x=\sqrt{\frac{6(0-3^2)}{-7500}}\\\\x=\sqrt{\frac{-54}{-7500}}\\\\x=0.085\ m =8.5\ cm\ \ [1\ m=100\ cm]$

Therefore, the maximum compression in the spring is 0.085 m or 8.5 cm.