Video: Determining the Ratio of the Change in the Extension of a Spring to the Change in Mass of the Load on the Spring

Ed Burdette

In building a house, carpenters use nails from a large box. The box is suspended from a spring twice during the day to measure the usage of nails. At the beginning of the day, the spring stretches 50 cm. At the end of the day, the spring stretches 30 cm What percentage of the nails have been used?

04:54

Video Transcript

In building a house, carpenters use nails from a large box. The box is suspended from a spring twice during the day to measure the usage of nails. At the beginning of the day, the spring stretches 50 centimetres. At the end of the day, the spring stretches 30 centimetres. What percentage of the nails have been used?

Let’s call that percentage of nails used capital 𝑁 sub 𝑒. As we begin our solution, let’s draw a diagram of the box suspended from the spring. In this diagram, on the far left, we have a picture of the spring hanging with no weight attached to it from the ceiling; this is the spring in its unstretched condition. Then at the beginning of the day, we attach a full box of nails to the end of the spring.

We’re told that the spring stretches a distance we’ve called π‘₯ sub one, where π‘₯ sub one is equal to 50 centimetres. Then the nails in the box are used up throughout the day. And at the end of the day, the box is once again hung from the spring, but this time because the mass of the box is less than it was in the morning, the spring is only stretched a smaller distance we’ve called π‘₯ sub two, where π‘₯ sub two is given as 30 centimetres.

We want to know the percentage of total nails used up throughout the day. Now one thing to notice about π‘š one and π‘š two as they hang from the spring is that these masses are in equilibrium; that is, they’re stationary as they hang. This means that for each mass, the spring force that pulls the mass up in order to restore the spring to its original length is balanced out by the force of gravity that exerts itself on these masses through the weight force. That’s why the masses π‘š one and π‘š two are stationary. So as an equation, we can write the spring force on mass one 𝐹 sub 𝑠 one is equal to mass one times the acceleration due to gravity 𝑔. Likewise, for the forces on π‘š two, 𝑓 sub 𝑠 two is equal to π‘š two times 𝑔.

At this point, let’s recall what the spring force is equal in terms of the spring constant π‘˜ and the spring’s displacement from equilibrium. The spring force 𝐹 sub 𝑠 is equal to negative π‘˜, the spring constant, times π‘₯, its displacement from equilibrium whether stretched or compressed. This means that in our force balance equations, we can replace 𝐹 sub 𝑠 one with negative π‘˜ times π‘₯ sub one and we can replace 𝐹 sub 𝑠 two with negative π‘˜ times π‘₯ sub two. The constants π‘˜ are the same because the spring that we’re working with is the same throughout the day.

Now let’s take the step of dividing these equations by one another. When we do that, we see some cancellation occur; 𝑔 the acceleration due to gravity cancels out as does negative π‘˜, the spring constant. We’re left with a simplified equation, which says the ratio of π‘₯ one to π‘₯ two is equal to the ratio of π‘š one to π‘š two. π‘š two is the mass of the box of nails at the end of the day and π‘š one is the mass of the box of nails at the beginning.

Let’s solve for π‘š two in terms of π‘š one. To do that, we cross multiply. π‘š two goes to the top left, π‘₯ one goes the opposite direction to the bottom right, and π‘₯ two goes to the top right. When the dust settles, we see that π‘š two is equal to the initial mass of the box of nails π‘š one multiplied by the fraction π‘₯ two divided by π‘₯ one. When we enter in the known values for π‘₯ two and π‘₯ one, 30 and 50 centimetres, respectively, we see that π‘š two is equal to π‘š one times 0.6.

In other words, π‘š two is 60 percent of π‘š one. This means that 𝑁 sub 𝑒, the percentage of nails that have been used up, is equal to one minus 0.6 times 100 percent. This equals 40 percent. Because the mass of the nails at the end of the day is 60 percent of the mass at the beginning, that means that 40 percent of the nails were used up throughout the day.

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