# A slingshot obeying Hooke's law is used to launch pebbles vertically into the air. You observe...

## Question:

A slingshot obeying Hooke's law is used to launch pebbles vertically into the air. You observe that if you pull a pebble back 2 cm against the elastic band the pebble goes 6 m high. Assuming that air drag is negligible, how high will the pebble go if you pull it back 4 cm? How far must you pull it back so it will reach 12 m? If you pull a pebble that is twice as heavy back 2 cm, how high will it go?

## Potential Energy under Hooke's Law

The following relation describes Hooke's law

$$\text{F}=-\text{k} x$$

• {eq}\rm{F} \equiv {/eq} force acting on the particle
• {eq}\rm{k}\equiv {/eq} spring constant
• {eq}x \equiv {/eq} displacement of the particle from a mean position

The potential energy of the particle can be obtained by integrating the force.

$$\text{U}=-\int_{0}^{x}\text{F} \text{d}x'=\int_{0}^{x}\text{k}x'=\dfrac{1}{2}\text{k}x^2$$

a. Potential energy due to gravity is given as

$$\text{U} =m\text{g}h$$

• {eq}m \equiv {/eq} mass of the particle
• {eq}\text{g} \equiv {/eq} acceleration due to gravity
• {eq}h \equiv {/eq} height

In our case, we can equate both the potential energies and notice that for the same pebble and the same slingshot,

$$\begin{array}{l} \dfrac{x_1^2}{h_1}=\dfrac{x_2^2}{h_2} \\ \ \\ \Rightarrow h_2=\left(\dfrac{x_2^2}{x_1^2 }\right) h_1 \\ \ \\ \Rightarrow \boxed{h_2 = \dfrac{4^2}{2^2}\times 6\ \text{m} = 24\ \text{m}} \end{array}$$

b. Also, we notice that

$$\begin{array}{l} x_2^2=\left(\dfrac{h_2}{h_1}\right)x_1^2 \\ \ \\ \Rightarrow x_2^2=\dfrac{12}{6}\times 2^2\ \text{cm}^2\\ \ \\ \Rightarrow \boxed{x_2= \sqrt{8}\approx 2.8\ \text{cm}} \end{array}$$

c. Now, originally, from equating the two potential energies, we got

$$\begin{array}{l} h=\left(\dfrac {\text{k}x^2}{2\text{g}}\right)\dfrac{1}{m} \end{array}$$

Therefore, if {eq}m {/eq} gets doubled, {eq}h {/eq} gets halved. Hence this time,

$$\boxed {h= 3\ \text{m}}$$ Hooke's Law & the Spring Constant: Definition & Equation

from

Chapter 4 / Lesson 19
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After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.