A 1.82 kg box moves back and forth on a horizontal frictionless surface between two different springs, as shown. The box is initially pressed against the stronger spring (k1 = 37.0 N/cm), compressing it 3.10 cm, and then is released from rest. By how much will the box compress the weaker spring (k2 = 18.5 N/cm)?
I tried using Uel=1/2 kx^2 to determine the potential force from spring1. And I used that force to spring2 again using U=1/2 kx^2 to determine x, but the answer was wrong.
U1=(1/2)(37.0)N/cm (3.10cm)^2
U2=(1/2)(18.5)N/cm (s cm)^2
U1=U2
2 * (3.10cm)^2 = (s cm)^2
19.22 cm^2 = s^2* cm^2
s = +- sqrt(19.22)
s = +- 4.38cm
To solve this problem, let's break it down step by step.
Step 1: Determine the potential energy stored in the stronger spring (spring 1) when it is compressed by 3.10 cm.
The formula to calculate the potential energy stored in a spring is Uel = 1/2 * k * x^2, where Uel is the elastic potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position.
Given that the spring constant for spring 1 is k1 = 37.0 N/cm and the compression of the spring is x1 = 3.10 cm, we can calculate the potential energy stored:
Uel1 = 1/2 * k1 * x1^2
= 1/2 * 37.0 N/cm * (3.10 cm)^2
= 1/2 * 37.0 N/cm * 9.61 cm^2
= 176.85 N·cm
Step 2: Determine the compression of the weaker spring (spring 2) when the box is released from rest.
Since the surface is frictionless and the box is initially at rest, the total mechanical energy of the system is conserved. Therefore, the potential energy stored in spring 1 will be transferred to spring 2 when the box is released.
Using the principle of conservation of energy, we can equate the potential energy stored in spring 1 to the potential energy stored in spring 2:
Uel1 = Uel2
Substituting the values we obtained earlier:
176.85 N·cm = 1/2 * k2 * x2^2
Solving for x2 (the compression of spring 2), we get:
x2^2 = 2 * (176.85 N·cm) / k2
x2 = √(2 * (176.85 N·cm) / k2)
x2 = √(2 * (176.85 N·cm) / 18.5 N/cm)
x2 = √(19.10 cm)
x2 ≈ 4.372 cm
Therefore, the box will compress the weaker spring (spring 2) by approximately 4.372 cm.