Question

A small block of mass 100 g is pressed against a horizontal spring fixed at one end to compress the spring through 5 cm (figure 8-E11). The spring constant is 100 N/m. When released, the block moves horizontally till it leaves the spring. Where will it hit the ground 2 m below the spring?

Figure

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=100g=0.1\mathrm{kg} \\ \mathrm{Compression}\mathrm{in}\mathrm{the}\mathrm{spring},x=5\mathrm{cm}=0.05\mathrm{m} \\ \mathrm{Spring}\mathrm{constant},k=100\mathrm{N}/\mathrm{m} \end{gathered}$$

Let v be the velocity of the block when it leaves the spring.

Applying the law of conservation of energy,

Elastic potential energy of the spring = Kinetic energy of the block

$$\begin{gathered} \frac{1}{2}m{\nu }^2=\frac{1}{2}k{x}^2 \\ \Rightarrow \nu =x\sqrt{\frac{k}{m}} \\ =\left(0.05\right)\times \sqrt{\frac{\left(100\right)}{0.1}} \\ =1.58\mathrm{m}/s \end{gathered}$$

For the projectile motion,

$$\begin{gathered} \mathrm{\theta }=0°,y=-2 \\ \mathrm{Now},y=\left(u·\sin \mathrm{\theta }\right)t-\frac{1}{2}g{t}^2 \\ -2=\left(-\frac{1}{2}\right)\times \left(9.8\right)\times t^2 \end{gathered}$$
$$\begin{gathered} \Rightarrow t=0.63\sec , \\ \mathrm{So},x=\left(u\cos \mathrm{\theta }\right)t \\ =\left(1.58\right)\times \left(0.36\right)=1\mathrm{m} \end{gathered}$$

Therefore, the block hits the ground at 1 m from the free end of the spring in the horizontal direction.