# Video: Conservation of Energy in Mechanical Energy Conversions

Figure 1 shows a metal block attached to the end of a spring. Initially, the metal block is held in place to counterbalance its weight, and the spring is in its unstretched equilibrium position. The mass of the metal block is 15 kg. The gravitational field strength is 9.8 m/s². Write down the equation that relates the gravitational potential energy of an object to its height above the ground, its mass, and the local gravitational field strength. The metal block is released from its initial position. Calculate the change in the gravitational potential energy of the block from its initial position to when it is 10 cm below its initial position. When the block is 10 cm below its initial position, 58% of its initial gravitational potential energy was transferred to its kinetic energy store. By how much has the block’s kinetic energy increase after falling 10 cm? Give your answer to two significant figures. Calculate the speed of the block after falling 10 cm. Give your answer to three significant figures. When the block reaches its lowest point, the potential energy stored in the spring is 35.28 J. Calculate the spring constant of the spring.

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### Video Transcript

Figure one shows a metal block attached to the end of a spring. Initially, the metal block is held in place to counterbalance its weight, and the spring is in its unstretched equilibrium position. The mass of the block is 15 kilograms. The gravitational field strength is 9.8 metres per second squared. Write down the equation that relates the gravitational potential energy of an object to its height above the ground, its mass, and the local gravitational field strength.

Since the equation we’re asked to write is a general one, we won’t need to reference any of the specific information about this particular system. We know that when it comes to gravitational potential energy, this is given as the product of three factors. It’s equal to an object’s mass multiplied by the local acceleration due to gravity multiplied by its height above some reference point.

We often see this equation written this way mathematically. And we can also express it using words. We can say that gravitational potential energy is equal to mass times acceleration due to gravity times height.

Now that we’ve written a general equation for gravitational potential energy, let’s apply this relationship to our scenario.

The metal block is released from its initial position. Calculate the change in gravitational potential energy of the block from its initial position to when it is 10 centimetres below its initial position.

To solve for this change, we can call the change in the height of the block given as 10 centimetres Δℎ. And we’ll also recall the mass of the block given as 15 kilograms and the acceleration due to gravity 9.8 metres per second squared.

To calculate the block’s change in gravitational potential energy, we can recall the equation we developed earlier. This equation tells us that the gravitational potential energy of an object is equal to its mass times the acceleration due to gravity times its height relative to some reference point.

In our case, we can say that the change in gravitational potential energy is equal to 𝑚 times 𝑔 times Δℎ. All these values are given to us. But before we plug them in, we want to convert the units on our change in height from centimetres into units of metres. This will help the units of that value agree with the units in 𝑔.

To help with this conversion, we can recall that 100 centimetres is equal to one metre. Therefore, 10 centimetres is equal to 0.10 metres. We’re now ready to plug in and solve for ΔPE sub 𝑔. When we do and we enter this expression on our calculator, we find a result to three significant figures of 14.7 joules. That’s the change in the gravitational potential energy of this block over this descent.

Next, let’s consider how the potential energy of the block connects with its kinetic energy.

When the block was 10 centimetres below its initial position, 58 percent of its initial gravitational potential energy was transferred to its kinetic energy store. By how much did the block’s kinetic energy increase after falling 10 centimetres? Give your answer to two significant figures.

We can say that the kinetic energy gained by the block is equal to the change in its gravitational potential energy multiplied by 58 percent or as a decimal 0.58. We calculated earlier that ΔPE sub 𝑔 is 14.7 joules. This number times 0.58 is equal to 8.5 joules. To two significant figures, that’s the kinetic energy gained by the block over this descent.

Next, let’s consider what this kinetic energy means in terms of the motion of the block.

Calculate the speed of the block after falling 10 centimetres. Give your answer to three significant figures.

We saw earlier that over this distance, the block’s change in kinetic energy was 8.5 joules. We can recall that the kinetic energy of a classical object is equal to one-half its mass times its speed squared. We can write then that ΔKE is equal to one-half 𝑚𝑣 squared, where 𝑣 is the speed.

When we rearrange to solve for speed by multiplying both sides by two over 𝑚 and taking the square root of both sides, we find that the speed of the block is equal to the square root of two times its change in kinetic energy divided by its mass.

We’ve calculated ΔKE and we know its mass from an earlier step. So we’re ready to plug in and solve for speed 𝑣. When we do plug in and enter this expression on our calculator, to three significant figures, we find a result of 1.06 metres per second. That’s the speed of the block after falling 10 centimetres.

Next, we’ll work with a type of potential energy we haven’t considered so far — spring potential energy.

When the block reaches its lowest point, the potential energy stored in the spring is 35.28 joules. Calculate the spring constant of the spring. Use the correct equation from the Physics Equation Sheet.

The potential energy stored in a spring, which we can call PE sub 𝑠, is equal to one-half times the spring constant 𝑘 multiplied by the spring’s displacement from equilibrium 𝑥 squared. In our question statement, we’re told the spring’s potential energy is 35.28 joules and we want to solve for 𝑘 based on this information.

We can start by algebraically rearranging this equation to solve for 𝑘. To do it, we’ll multiply both sides by two divided by 𝑥 squared. The spring constant 𝑘 then is equal to two times PE sub 𝑠 over 𝑥 squared.

We’re given the potential energy of the spring and we’re told that 𝑥, this distance from equilibrium, occurs when the block reaches its lowest point. When we consult Figure one, we see that this lowest point occurs when the spring is extended by 24 centimetres. This tells us that 𝑥 is 24 centimetres.

But before we plug in this value for 𝑥, we want to convert it into units of metres. Recalling that 100 centimetres is equal to one metre, this means that 24 centimetres is equal to 0.24 metres. When we plug in this value for 𝑥 as well as the given value for PE sub 𝑠, to four significant figures, we find that 𝑘 is 1225 newtons per metre. That’s the spring constant of this spring.