Solving Hooke's Law Problem: 2kg + 3kg Masses, k=136N/m, F=18N

[darklich21]

Homework Statement

A 2kg mass and a 3kg mass are on a horizontal frictionless surface, connected by a massless spring with spring constant k=136N/m. An 18 N force is applied to the larger mass. How much does the spring stretch from its equilibirum length? (answer in cm)

Homework Equations

F=-kd
k= 136N/m
F=18N

The Attempt at a Solution

I did the obvious, I used the equation above and solved for D: 18/136 = 0.132m x 100= 13.2 cm, which is wrong. I obviously neglected the two masses, which is probably where I went wrong? Can anyone help me out by providing the proper equation and solution?

 

[noblegas]

Did you try drawing a free body diagram of the two mass strings connected by a massless string? It might be easier for you to write down your equations for this problem after you draw the free body diagram. Do you agree that the spring force and the applied force make up the net force?

 

[tomcue]

As a scientist, the first step in solving this problem would be to draw a free-body diagram of the system. This will help us visualize the forces acting on the masses and the spring. From the given information, we can see that the 18N force is applied to the larger mass, while the smaller mass is connected to the spring. The spring itself exerts a restoring force on both masses, which is given by Hooke's Law: F=-kx, where F is the force applied by the spring, k is the spring constant, and x is the displacement from the equilibrium position.

Next, we can use the principle of conservation of energy to solve for the displacement of the spring. The total energy in the system is equal to the sum of the potential energy stored in the spring and the kinetic energy of the masses. At equilibrium, the potential energy is zero and all the energy is in the form of kinetic energy. When the 18N force is applied, the system will start to oscillate, with the potential energy stored in the spring increasing and the kinetic energy decreasing. At the point where the kinetic energy is zero, the potential energy of the spring will be equal to the work done by the 18N force. This can be represented by the equation: 1/2 kx^2 = 18 x d, where d is the displacement of the larger mass.

We can now solve for x, the displacement of the spring, by substituting the given values into the equation: x=√(2F/k) = √(2x18/136) = 0.243m = 24.3cm.

Therefore, the spring would stretch by 24.3cm from its equilibrium length.

It is important to note that this solution assumes that the masses are connected by a massless spring and that there is no friction present. In real-world situations, these assumptions may not hold and may affect the accuracy of the solution. It is also important to check the units of the given values and make sure they are consistent before plugging them into equations.