Question Video: Mechanical Energy Conversion

A mobile phone with a mass of 120 g is held in the hand of a tourist relaxing on a boat on holiday. The tourist distractedly holds the phone just over the surface of the water and accidentally loses hold of it. The phone sinks through the water, and the instantaneous speed at which it sinks when at a depth of 10 cm is 1.1 m/s. How much work is done moving water out of the phone’s path as it sinks by 10 cm?

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Video Transcript

A mobile phone with a mass of 120 grams is held in the hand of a tourist relaxing on a boat on holiday. The tourist distractedly holds the phone just over the surface of the water and accidentally loses hold of it. The phone sinks through the water. And the instantaneous speed at which it sinks when at a depth of 10 centimetres is 1.1 metres per second. How much work is done moving water out of the phone’s path as it sinks by 10 centimetres?

Okay, so to answer this question, let’s first start off by drawing a diagram. So here’s the surface of the water that’s just been disturbed by the poor tourist’s mobile phone falling through the water. Now we know that the distance that the mobile phone falls is 10 centimetres. And we also know that when the phone is at a depth of 10 centimetres, it has an instantaneous velocity of 1.1 metres per second.

Now instantaneous velocity, basically just means the velocity of the phone at that instant, in other words, the velocity of the phone when it’s 10 centimetres below the surface. Finally, we also know that the phone has a mass of 120 grams. What we’ve been asked to do is to find out the amount of work done, which we’ll call 𝑊, in moving water out of the phone’s path.

Before we start working anything out though is to name all of the quantities that we’ve labelled in the diagram, starting with the 10-centimetre distance that the phone has fallen. Let’s name it Jeffrey. Let’s name it Chris. No, obviously I’m kidding. We’re gonna have to give it a mathematical name. And the letter 𝑑 is probably a good name because this is the depth that the phone has fallen.

Secondly, we can also label the mass of the phone, which we’ll call 𝑚, and finally the velocity of the phone at that instant, which we’ll call 𝑉. Finally, it’s important to note that 𝑊 is the work done in moving water out of the phone’s path. In other words, this is the work done by the phone on the water.

So to go about finding the solution to this problem, let’s first erase the question so we’ve got more space to work. Now the first thing we can do is to recall that work done on an object, which we’ll call 𝑊, is given by the force on the object multiplied by the distance that the object travels. So in order for us to work out the work done moving water out of the phone’s path, we need to work out the force exerted by the phone on the water multiplied by the distance moved.

Now the distance travelled is fairly simple. It’s 10 centimetres. We’ve been given this distance already. But what we don’t know is the force exerted. Specifically, because we’re trying to work out the work done by the phone on the water, we need to work out the force exerted by the phone on the water.

Now to do this, what we need to use is known as Newton’s third law of motion. What this law tells us is that if an object — let’s call it object one — exerts a force on a second object, object two, then object two exerts an equal and opposite force on object one.

Now why is this useful? Well, let’s first consider the forces acting on the phone. So here is the phone falling through the water. Now there are two forces acting on the phone. Firstly, we’ve got the gravitational force pulling it downwards. And we’ll call this force 𝐹 sub grav, because it’s the gravitational force.

Now the other force acting on the phone is the drag force. This force is exerted by the water on the phone. And it opposes the motion of the phone. In other words, it’s trying to slow the phone down. And hence it acts in an upward direction. Now we’ll call this force 𝐹 sub drag.

And as we’ve already said, 𝐹 sub drag is the force exerted by the water on the phone. So using Newton’s third law, we can therefore say that the force exerted by the phone on the water has the same magnitude or size as 𝐹 sub drag. Of course, it acts in the opposite direction. But that’s not relevant to us now.

It does make sense though. The phone is being pulled downwards by gravity. And hence, as the phone moves downward, it will exert a downward force on the water. But the size of this force is what’s important. And the size of this force is equal to 𝐹 sub drag. And by the way, that’s the force that we’re looking for here when we try to work out the work done by the phone on the water. So in simple terms, we’re trying to find out what 𝐹 sub drag is.

Let’s go about doing that by first working out the net force on the phone. The idea of that is that there are two forces acting on the phone as we’ve already seen: 𝐹 sub grav and 𝐹 sub drag. The net result of this force will result in an acceleration of the phone. So how do we know this?

Well, we know this by using yet another one of Newton’s laws of motion, this time the second law. Now this law tells us that the net force on an object, which we’ll call 𝐹 sub net, is equal to the mass of that object multiplied by its acceleration. So what is 𝐹 sub net? What is the net force on the phone?

Well, the net force is simply the two forces acting on the phone added together, whilst also accounting for the direction in which they act. In other words, the net force is equal to 𝐹 sub grav minus 𝐹 sub drag. When we’ve written this equation down, we’ve implicitly assumed that the downward direction is positive and, therefore, the upper direction is negative. Hence, the downward-acting force, 𝐹 sub grav, is positive. And we’ve put a negative sign in front of the upward-acting force. But that is the expression for 𝐹 sub net, the net force on the phone. So let’s first of all work out an expression for 𝐹 sub grav.

𝐹 sub grav is simply the gravitational force on the phone. Or, in other words, it’s the weight of the phone. Now we can recall that the weight of an object, 𝐹 sub grav, is given by multiplying the mass of the object by the gravitational field strength of the Earth.

Now realize that we haven’t used 𝑊 for weight. That’s because we’ve already used 𝑊 to represent the work done by the phone on the water. So let’s just stick with 𝐹 sub grav instead. Now luckily for us, we know the mass of the phone. We’ve been told this in the question, which is 120 grams. And we can recall that the gravitational field strength of the Earth is 9.8 metres per second squared.

Therefore, we have an expression for 𝐹 sub grav. So let’s replace that in our equation here. And then we can move on to looking at what 𝐹 sub net is. Now as we’ve said already, 𝐹 sub net is given by multiplying the mass of the object by the acceleration of the object. We don’t yet know the acceleration of the object. But we can work this out. This is because we know firstly that the initial velocity of the phone has to be zero metres per second because the phone was initially held at the surface of the water by the tourist. And so it was held stationary just as it was dropped. And we’ll call this initial velocity 𝑢.

Secondly, we know that the phone travels downwards a distance 𝑑, which happens to be 10 centimetres. Thirdly, we also know that when the phone has travelled 10 centimetres downward, it has a velocity which we’ve called 𝑉 of 1.1 metres per second. And using these three pieces of information, we can work out the acceleration of the phone.

We can do this by using one of the SUVAT equations. Specifically, the one we want tells us that the final velocity squared is equal to the initial velocity squared plus two times the acceleration times the distance travelled. And we can rearrange this equation to find out what 𝑎 is. To do this, we first subtract 𝑢 squared from both sides of the equation. This way, 𝑢 squared cancels on the right-hand side, leaving us with just two 𝑎𝑑. Then we can divide both sides of the equation by two 𝑑 so that the two and the 𝑑 on the right-hand side cancel out. And now we have an expression for 𝑎 in terms of 𝑉, 𝑢, and 𝑑.

Now why were we trying to work out 𝑎 in the first place? Oh yeah! That’s right, so that we could work out what 𝐹 sub net is. We multiply this acceleration that we’ve just found by the mass of the phone in order to give us the net force on the phone. In other words, we multiply both sides of this equation by 𝑚, the mass of the phone. And we cannot forget to put parentheses around 𝑉 squared minus 𝑢 squared. And this whole thing is equal to 𝐹 sub net.

What this means is that we can take this expression and plug it into 𝐹 sub net. Doing this gives us 𝑚 multiplied by 𝑉 squared minus 𝑢 squared over two 𝑑 is equal to 𝑚𝑔 minus 𝐹 sub drag. Now as we said earlier, 𝐹 sub drag is what we’re trying to find out because that’s this force here. So let’s rearrange to find what 𝐹 sub drag is.

Firstly, we can add 𝐹 sub drag to both sides of the equation. This way, 𝐹 sub drag cancels on the right-hand side. Then we can subtract this whole term from both sides of the equation so that we’re left with just 𝐹 sub drag on the left-hand side. Now remember, 𝐹 sub drag is the drag force on the phone. In other words, it’s the force exerted by the water on the phone. And as we said earlier, using Newton’s third law of motion, the force exerted by the phone on the water is going to have the same magnitude. It’s going to have the same size.

Hence, if we’re just worrying about the sizes of the forces and not the direction in which they’re acting, we can say that the work done by the phone on the water, which is what we’re trying to find out eventually, is given by 𝑊 is equal to 𝐹 sub drag multiplied by the distance 𝑑. In other words, what we need to do to find our final answer is to multiply both sides of this equation by 𝑑.

Now, of course, once again, we cannot forget the parentheses around the entire expression. And of course, the left-hand side simply becomes the work done 𝑊. And on the right-hand side, we can distribute this 𝑑 to both terms in the parentheses. What that looks like is 𝑚𝑔𝑑 minus 𝑚 multiplied by 𝑉 squared minus 𝑢 squared over two, because remember we had a 𝑑 here and we multiplied this term by 𝑑 as well, so the 𝑑s cancelled.

Now at this point, we have an expression for the work done moving the water out of the phone’s path. In other words, we have an expression for the work done by the phone on the water. And we have it in terms of quantities that we already know, because we already know the mass of the phone. We already know the gravitational field strength of the Earth. We know the distance travelled by the phone in the water. We also know the final velocity of the phone and the initial velocity of the phone. So at this point, all that’s left for us to do is to plug stuff in.

But before we do that, we need to convert all of the quantities we’ve been given to standard units. Let’s start with the mass of the phone 𝑚. We know that 𝑚, the mass, is equal to 120 grams. However, the standard unit of mass is kilograms. So we need to convert this mass into kilograms. To do this, we can recall that one kilogram is equal to 1000 grams. We can divide both sides of this equation by 1000 so that we find that, on the right-hand side, the 1000s cancel. And what we’ve got left is one gram.

And so as we’ve seen, the right-hand side simplifies to one gram. And the left-hand side simplifies to one thousandth of a kilogram. But we’re not trying to work out what one gram is in kilograms. We’re trying to work out what 120 grams is. So to do this, we multiply both sides of the equation by 120. And so the right-hand side leaves us with 120 grams. And on the left, we’re left with 0.12 kilograms. Now that’s the mass of the phone in standard units. So we replace the mass 𝑚 here with 0.12 kilograms.

Next, what we can do is to look at the values of 𝑢, 𝑑, 𝑉, and 𝑎. Now 𝑢 is given in metres per second. That is the standard unit of the velocity. So we don’t need to convert it.

𝑑, however, is given in centimetres, whereas the standard unit of a distance is metres. So we need to convert 𝑑 to metres. To do this, we’ll recall that one metre is equal to 100 centimetres. Now if we divide both sides of this equation by 10, we find that the right-hand side becomes 10 centimetres and the left-hand side becomes 0.1 metres. Therefore, we’ve replaced the value of 𝑑 with 0.1 metres. And at this point, 𝑑 is now in the correct units of metres.

Now thirdly, the final velocity of the phone 𝑉 is given in metres per second. So once again, this is correct. And fourthly, the units of acceleration, well we haven’t calculated the acceleration yet. But we used the SUVAT equations and these three quantities to work out what the acceleration was. And then we used the expression for acceleration in this final term here. But if we use these three quantities in the correct units 𝑢, 𝑑, and 𝑉, then the acceleration will also come out in its standard unit of metres per second squared. So we don’t need to worry about this one.

And it’s at this point that we can plug in all the values into our final expression here. So the work done 𝑊 by the phone on the water is equal to, firstly, the mass of the phone, which is 0.12 kilograms, multiplied by the gravitational field strength of the Earth, which is 9.8 metres per second squared, multiplied by the distance travelled by the phone, which we said was 0.1 metres, minus the mass of the phone once again, which is 0.12 kilograms, multiplied by the final velocity of the phone squared, which is 1.1 metres per second all squared, minus the initial velocity of the phone squared, which is zero squared. And the whole thing is divided by two.

Now we can evaluate this expression to finally give us a value for the work done by the phone on the water. And because we’ve been careful to put everything in standard units, the work done is also going to be in its standard unit, which is joules. So evaluating this expression on the right-hand side, we find that the work done in moving the water out of the phone’s path is 0.045 joules. And at this point, we have the final answer to our question.

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