Question

A slingshot has a spring constant of 550 N/m. A 40 gram pellet is placed in the slingshot and then drawn back 20 cm from equilibrium. If the pellet is released straight up, how far up will the pellet travel from its starting position?

Answer 1

28.06 metres

Step-by-step explanation

Step 1

find the elastic potential energy

it is given by:

`\begin{gathered} E_k=(1)/(2)kx^2 \\ \text{where kis the constant of the slingshot} \\ x\text{ is the distance} \end{gathered}`

so

convert the measure into meteres and kg

`\begin{gathered} 40\text{ gr =}\frac{\text{40}}{1000}kg\text{ = 0.040 }kg \\ 20cm=\text{ }\frac{\text{20}}{100}=0.2m \end{gathered}`

so

`\begin{gathered} E_k=(1)/(2)kx^2 \\ E_k=(1)/(2)\cdot550(N)/(m)\cdot(0.2)^2 \\ E_k=(1)/(2)\cdot550(N)/(m)\cdot(0.2)^2 \\ E_k=11\text{ J} \end{gathered}`

Step 2

now, use the conservation of energy law to find the maximum heigth

`\begin{gathered} \text{elastic energy = potential eneregy} \\ 11\text{ J = mgh} \\ so \\ h=(11)/(mg) \\ \text{replace} \\ h=\frac{11\text{ J}}{0.040kg\cdot9.8(m)/(s^2)} \\ h=28.06\text{ meters} \end{gathered}`

therefore, the answer is

28.06 metres

I hope this helps you