Question

a student is about to launch a 0.1 kg rock from a slingshot as shown. the natural length of the spring is shown in the insert. the initial time is defined as just before the launch when the student has pulled the rock backwards and is holding the rock at rest with a 86 n force (note that the natural and stretched total band lengths are shown). the final time is defined immediately after the launch. neglect loss.
how much did the student change the length of the bands when she pulled the rock backwards?
unit
what is the effective spring constant of the slingshot bands?
unit
list the energy types at the initial and final time and whether work and loss (due to non - conservative forces) occur as well as the corresponding amounts of energy.
initial:
unit
work?
unit
loss?
unit
final:
unit

Explanation:

Step1: Determine the stretch of the spring

The natural length of the spring (initial length when no force is applied) is \( l_0 = 0.16 \, \text{m} \), and the stretched length (final length when the force is applied) is \( l = 0.51 \, \text{m} \). The stretch \( x \) of the spring is the difference between the stretched length and the natural length:
\( x = l - l_0 = 0.51 \, \text{m} - 0.16 \, \text{m} = 0.35 \, \text{m} \).

Step2: Calculate the spring constant using Hooke's Law

Hooke's Law states that the force \( F \) exerted by a spring is \( F = kx \), where \( k \) is the spring constant and \( x \) is the stretch. We know the force \( F = 86 \, \text{N} \) and the stretch \( x = 0.35 \, \text{m} \). Rearranging Hooke's Law to solve for \( k \):
\( k = \frac{F}{x} \).

Substitute the known values:
\( k = \frac{86 \, \text{N}}{0.35 \, \text{m}} \approx 245.71 \, \text{N/m} \).

Answer:

The effective spring constant of the slingshot bands is approximately \( \boldsymbol{246 \, \text{N/m}} \) (or \( 245.7 \, \text{N/m} \) with more precision).