Question

A shown in figure there is a spring block system. Block of mass $500$ g is pressed against a horizontal spring fixed at one end to compress the spring through $5.0$ cm. The spring constant is $500$ N/m. When released, the block moves horizontally till it leaves the spring. Calculate the distance where it will hit the ground $4$ m below the spring?

• A. $6m$
• B. $4m$
• C. $8m$
• D. $\sqrt{2}$m

Answer 1

The correct option is D $\sqrt{2}$m

Mass of the block $m=0.5kg$

When block released, the block moves horizontally with speed $v$ till it leaves the spring.

By energy conservation, we get $\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2$

where $x=0.05m$ is the compression of the spring, $k=500N/m$ is the spring constant.

$\therefore$ We get $v^2=\frac{k{x}^2}{m}\Rightarrow v=\sqrt{\frac{k{x}^2}{m}}$

Time of flight $t=\sqrt{\frac{2H}{g}}$

where $H=4m$

So, horizontal distance traveled from the free end of the spring $R=v\times t$

$\therefore$ $R=\sqrt{\frac{k{x}^2}{m}}\times \sqrt{\frac{2H}{g}}$

Or $R=\sqrt{\frac{500\times (0.05{)}^2}{0.5}}\times \sqrt{\frac{2\times 4}{10}}=\sqrt{2}m$

So, the block will hit the ground at a horizontal distance of $\sqrt{2}m$ from the free end of the spring.