A 2kg block is dropped from a height of 0.4 m on a spring of force constant K = 1960 Nm⁻¹. The maximum compression of the spring is

Correct Answer: Option 0.1m

Step-by-step Explanation:

To find the maximum compression of the spring, we can apply the Law of Conservation of Energy. When the block is dropped, it loses gravitational potential energy and the spring gains elastic potential energy.

Let:
Mass of the block, $m=2\mathrm{kg}$
Initial height above the spring, $h=0.4m$
Spring constant, $K=1960N{m}^{-1}$
Acceleration due to gravity, $g=9.8m{s}^{-2}$
Maximum compression of the spring, $x$

As the block falls and compresses the spring by a maximum distance $x$, the total vertical height it descends is $(h+x)$.

At the point of maximum compression, the block momentarily comes to rest, meaning its kinetic energy is zero. Therefore, the total loss in gravitational potential energy of the block is completely converted into the elastic potential energy stored in the spring:

$mg(h+x)=\frac{1}{2}K{x}^2$

Substituting the given values into the equation:

$2\times 9.8\times (0.4+x)=\frac{1}{2}\times 1960\times x^2$

Simplifying both sides:

$19.6(0.4+x)=980x^2$

Divide both sides of the equation by 19.6:

$0.4+x=50x^2$

Rearranging this into a standard quadratic equation format:

$50x^2-x-0.4=0$

To eliminate the decimal, multiply the entire equation by 10:

$500x^2-10x-4=0$

Dividing the equation by 2:

$250x^2-5x-2=0$

Now, we factor the quadratic equation by splitting the middle term:

$250x^2-25x+20x-2=0$

$25x(10x-1)+2(10x-1)=0$

$(25x+2)(10x-1)=0$

This gives two possible solutions for $x$:

$25x+2=0\Rightarrow x=-0.08m$

or

$10x-1=0\Rightarrow x=0.1m$

Since the compression $x$ must be a positive distance, we discard the negative root.

Thus, the maximum compression of the spring is 0.1 m.