Question Video: Conservative Forces and Potential Energy | Nagwa Question Video: Conservative Forces and Potential Energy | Nagwa

Question Video: Conservative Forces and Potential Energy

Two 8.00 kg masses are connected to each other by a spring with a force constant of 25.0 N/m and a rest length of 1.00 m. The spring has been compressed to 0.900 m in length. The masses travel toward each other, and each mass moves at 0.500 m/s. What is the total energy in the system?

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Video Transcript

Two 8.00-kilogram masses are connected to each other by a spring with a force constant of 25.0 newtons per meter and a rest length of 1.00 meters. The spring has been compressed to 0.900 meters in length. The masses travel toward each other. And each mass moves at 0.500 meters per second. What is the total energy in the system?

We can label the total system energy we want to solve for capital 𝐸 and begin by drawing a sketch of the situation. In this scenario, we have two identical masses β€” we’ve labeled them π‘š β€” of value 8.00 kilograms, connected by a spring of spring constant π‘˜. The spring started out at a length we’ve called 𝐿 sub 𝑖 for its initial length 1.00 meters. And now it’s at a length we’ve called 𝐿 sub 𝑓, 0.900 meters. The two masses move towards one another at speeds 𝑣, where 𝑣 is given as 0.500 meters per second. Knowing all this, we want to solve for the total energy in the system at this moment in time.

We know that this total energy will be equal to the sum of the system’s kinetic plus its potential energy. We recall that kinetic energy, in general, is equal to one-half an object’s mass times its speed squared. And since we have two identical masses, π‘š, moving at identical speeds, 𝑣, we can rewrite this expression for kinetic energy simply as π‘š times 𝑣 squared.

When it comes to the potential energy of our system, we have a spring which has been compressed from its natural length. The potential energy due to an elastic deformation is equal to one-half a spring’s constant, π‘˜, multiplied by its displacement from equilibrium, Ξ”π‘₯, squared. In our case, we can write this elastic potential energy as one-half the force constant, π‘˜, times the quantity 𝐿 sub 𝑖 minus 𝐿 sub 𝑓 squared.

In the problem statement, we’re given π‘š, 𝑣, π‘˜, 𝐿 sub 𝑖, and 𝐿 sub 𝑓. So we’re ready to plug in and solve for 𝐸. When we enter in the values for π‘š, 𝑣, π‘˜, 𝐿 sub 𝑖, and 𝐿 sub 𝑓 and enter this expression on our calculator, we find that, to three significant figures, 𝐸 is 2.13 joules. That’s the total energy of the system, kinetic plus potential.

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