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 0.9 m, we can use the formula for the potential energy stored in a spring:

Potential energy (U) = (1/2) k x^2

where k is the force constant of the spring and x is the displacement from the equilibrium position.

Given:
Mass (m) = 0.44 kg
Displacement (x) = 0.9 m
Force constant (k) = 2 N/m

Let's plug in the values into the formula:

U = (1/2) k x^2
U = (1/2) * 2 N/m * (0.9 m)^2
U = (1/2) * 2 N/m * 0.81 m^2
U = 0.81 N*m = 0.81 J

Therefore, the potential energy of the system when the spring is stretched 0.9 m is 0.81 Joules (J).

To find the potential energy of the system when the spring is stretched 0.9 m, we can use the formula for the potential energy of a spring:

Potential Energy (PE) = (1/2) * k * x^2

where k is the force constant of the spring and x is the displacement of the mass from its equilibrium position.

Given that the force constant (k) is 2 N/m and the displacement (x) is 0.9 m, we can substitute these values into the formula to calculate the potential energy.

PE = (1/2) * 2 N/m * (0.9 m)^2
= (1/2) * 2 N/m * 0.81 m^2
= 0.81 J

Therefore, the potential energy of the system when the spring is stretched 0.9 m is 0.81 J.