A 0.44 kg mass at the end of a horizontal spring is displaced 1.5 m and released, then moves in SHM at the end of the spring of force constant 2 N/m.
Find the potential energy of the system when the spring is stretched 0.9 m. The
acceleration of gravity is 9.8 m/s2 . Answer in units of J.
To find the potential energy of the system when the spring is stretched by 0.9 m, we need to use the equation for potential energy in a spring:
Potential Energy (U) = (1/2) * k * x²
Where:
- U is the potential energy
- k is the force constant of the spring
- x is the displacement of the spring from its equilibrium position
Given:
- Mass (m) = 0.44 kg
- Displacement (x) = 0.9 m
- Force constant (k) = 2 N/m
First, we need to calculate the spring constant using Hooke's Law:
k = m * ω²
Where:
- m is the mass
- ω is the angular frequency of the harmonic motion
To find ω, we can use the formula:
ω = √(k / m)
Let's calculate ω:
ω = √(2 N/m / 0.44 kg)
= √(4.55 N/kg)
≈ 2.13 rad/s
Now that we have the angular frequency, we can calculate the potential energy using the formula mentioned above:
U = (1/2) * k * x²
U = (1/2) * 2 N/m * (0.9 m)²
= (1/2) * 2 N/m * 0.81 m²
= 0.81 J
Therefore, the potential energy of the system when the spring is stretched by 0.9 m is approximately 0.81 J.