# Question Video: Using the Principle of Conservation of Energy to Determine the Energy Dissipated by Friction of an Object That Compresses a Spring

A box that has 50 J of kinetic energy slides on a frictionless surface. The box hits a spring and compresses the spring a distance of 25 cm from its equilibrium length. If the same box with the same initial energy slides on a rough surface, it only compresses the spring a distance of 15 cm. How much energy must have been lost by the box when sliding on the rough surface?

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### Video Transcript

A box that has 50 joules of kinetic energy slides on a frictionless surface. The box hits a spring and compresses the spring a distance of 25 centimeters from its equilibrium length. If the same box with the same initial energy slides on a rough surface, it only compresses the spring a distance of 15 centimeters. How much energy must have been lost by the box when sliding on the rough surface?

We’ll call this lost energy 𝐸 sub 𝐿. Let’s begin our solution by drawing a diagram of the situation. In the first picture of the sliding box, the box slides across a frictionless surface with energy 𝐸 and then compresses a spring a distance we’re given; we’ll call that 𝑥 sub one, and that value is 25 centimeters.

In the second scenario, the same box approaches the spring with the same amount of initial energy, but this time, the box is not moving over a frictionless surface, but over a rough surface that takes energy away due to friction. This energy loss results in the fact that the box only compresses the spring a distance now of 𝑥 sub two, which equals 15 centimeters.

Given that 𝐸, the initial energy of the box, is 50 joules, we want to solve for the energy lost to friction on the rough surface, 𝐸 sub 𝐿. To begin, let’s recall the equation for the potential energy stored up in a spring.

Spring potential energy is equal to one-half 𝑘, the spring constant, multiplied by the spring’s displacement from equilibrium, 𝑥 squared. In the case of the box sliding along the frictionless surface, we can write that 𝐸, the initial energy of the box, is losslessly converted into spring potential energy.

We know all the values in this equation except for 𝑘, the spring constant. We can rearrange to solve for 𝑘. We do that by multiplying both sides by two divided by 𝑥 sub one squared, which cancels out the one-half with the two and 𝑥 sub one squared terms on the right side of the equation.

So spring constant 𝑘 is equal to two times 𝐸 divided by 𝑥 sub one squared. 𝐸 is given as 50 joules and 𝑥 sub one is 25 centimeters or 0.25 meters. When we multiply these values through, we find that 𝑘 is equal to 1600 newtons per meter.

Knowing the spring constant 𝑘, we can now solve for the energy loss due to friction in the second case, where the box slid along the rough surface. That loss, 𝐸 sub 𝐿, is equal to the difference between the spring potential energy in case one and the spring potential energy in case two.

We can rewrite this as one-half 𝑘 times 𝑥 sub one squared minus one-half 𝑘 times 𝑥 sub two squared. We can factor out one-half 𝑘 from the right side of the equation, and we can then plug in the values for 𝑘, 𝑥 sub one, and 𝑥 sub two.

𝑘 is 1600 newtons per meter, 𝑥 sub one in units of meters is 0.25, and 𝑥 sub two in units of meters is 0.15. When we multiply these values through, we find a result of 32 joules. That’s how much energy was lost to friction between the box and the rough surface in case two.

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