A rectangular block of mass $5 \mathrm{~kg}$ attached to a horizontal spiral spring executes simple harmonic motion of amplitude $1 \mathrm{~m}$ and time period $3.14 \mathrm{~s}$. The maximum force exerted by spring on block is _________ N

<p>To find the maximum force exerted by the spring on the block, we can use Hooke&#39;s law and the properties of simple harmonic motion.</p> <p>First, let&#39;s find the angular frequency $\omega$:</p> <p>$\omega = \frac{2\pi}{T}$</p> <p>where $T = 3.14\,\mathrm{s}$ is the time period.</p> <p>$\omega = \frac{2\pi}{3.14} \approx 2\,\mathrm{rad/s}$</p> <p>Now, let&#39;s find the maximum velocity $v_{max}$ of the block:</p> <p>$v_{max} = \omega A$</p> <p>where $A = 1\,\mathrm{m}$ is the amplitude.</p> <p>$v_{max} = 2\,\mathrm{rad/s} \times 1\,\mathrm{m} = 2\,\mathrm{m/s}$</p> <p>Next, we can find the spring constant $k$ using the mass of the block $m = 5\,\mathrm{kg}$ and the angular frequency $\omega$:</p> <p>$\omega^2 = \frac{k}{m} \Rightarrow k = m\omega^2$</p> <p>$k = 5\,\mathrm{kg} \times (2\,\mathrm{rad/s})^2 = 20\,\mathrm{N/m}$</p> <p>Finally, we can find the maximum force exerted by the spring on the block. At maximum displacement, the force is given by Hooke&#39;s law:</p> <p>$F_{max} = kA$</p> <p>$F_{max} = 20\,\mathrm{N/m} \times 1\,\mathrm{m} = 20\,\mathrm{N}$</p> <p>The maximum force exerted by the spring on the block is $20\,\mathrm{N}$.</p>