A mass of \(1 \, kg\) falls from a height of \(1 \, m\) and lands on a massless platform supported by a spring having spring constant \(15 \, Nm^{-1}\) as shown in the figure. The maximum compression of the spring is
\[
(\text{acceleration due to gravity} = 10 \, m/s^2)
\]
Show Hint
In spring compression problems, include the entire downward displacement of the body while calculating the loss in gravitational potential energy.
Step 1: Understand the energy transformation.
When the \(1 \, kg\) mass falls, its gravitational potential energy gets converted into spring potential energy at maximum compression.
At maximum compression, the velocity becomes zero.
Let the maximum compression of the spring be \(x\).
Step 2: Calculate the total loss in gravitational potential energy.
The block first falls through \(1 \, m\) before touching the spring and then further compresses the spring by \(x\).
Therefore, total distance fallen by the block is
\[
1+x
\]
Hence, total loss in gravitational potential energy is
\[
mg(1+x)
\]
Substituting values,
\[
1 \times 10 \times (1+x)
\]
\[
=10(1+x)
\]
Step 3: Write the spring potential energy.
At maximum compression, energy stored in the spring is
\[
\frac12 kx^2
\]
Given,
\[
k=15 \, Nm^{-1}
\]
So, spring energy becomes
\[
\frac12 \times 15 \times x^2
\]
\[
=\frac{15}{2}x^2
\]
Step 4: Apply conservation of energy.
\[
10(1+x)=\frac{15}{2}x^2
\]
Multiply both sides by \(2\),
\[
20(1+x)=15x^2
\]
\[
20+20x=15x^2
\]
\[
15x^2-20x-20=0
\]
Divide by \(5\),
\[
3x^2-4x-4=0
\]
Step 5: Solve the quadratic equation.
Using the quadratic formula,
\[
x=\frac{4\pm\sqrt{(-4)^2-4(3)(-4)}}{2(3)}
\]
\[
x=\frac{4\pm\sqrt{16+48}}{6}
\]
\[
x=\frac{4\pm\sqrt{64}}{6}
\]
\[
x=\frac{4\pm8}{6}
\]
Possible values are
\[
x=2 \quad \text{or} \quad x=-\frac23
\]
Since compression cannot be negative,
\[
x=2 \, m
\]
Step 6: Final conclusion.
Hence, the maximum compression of the spring is
\[
\boxed{2 \, m}
\]
