Question Video: Mechanical Energy Conversion Physics • 9th Grade

An object with a velocity 𝑣 is slowed to rest by a constant force 𝐹 over a distance of 12 m. If the object’s velocity is increased to 5𝑣 and then 𝐹 is again applied to decelerate the object, how far does the object move between the force being applied and the object coming to rest?

07:30

Video Transcript

An object with a velocity 𝑣 is slowed to rest by a constant force 𝐹 over a distance of 12 meters. If the object’s velocity is increased to five 𝑣 and then 𝐹 is again applied to decelerate the object, how far does the object move between the force being applied and the object coming to rest?

Okay, so in this question, we know that we’ve got an object which initially has a velocity 𝑣, which is then slowed to rest by a constant force 𝐹. And it slows to rest over a distance of 12 meters. If we then take that same object and change its initial velocity to five 𝑣 and then use the same force 𝐹 to decelerate the object, we need to find out how far the object moves between the force being applied and the object coming to rest.

So basically, we’ve got two scenarios where the object that we’ve got is decelerating. The first scenario is where its initial velocity is 𝑣 and the second scenario is when its initial velocity is five 𝑣. Let’s draw a diagram of both scenarios.

First of all, let’s say that this is our object — just a regular ball with a mass 𝑚. We don’t actually know what this mass is, but as we’ll see later we don’t need to know. Now, in the first scenario, this object is starting out with a velocity 𝑣. So let’s say it’s moving towards the right. And we know it travels some distance and is decelerating the whole time until it stops. The distance it travels while decelerating until it stops has been given as 12 meters. Let’s call this distance 𝑠. Now what’s slowing the object down is a force that we apply and this force must be in the opposite direction because it’s decelerating the object. It’s slowing it down. So that’s the first situation.

Now, let’s draw the second scenario. This time we’ve got the same object again. This ball with a mass of 𝑚. But it starts out with a velocity of five 𝑣. And once again, we apply the same force to it 𝐹 so that it decelerates over some distance until it stops. Now, we’re asked to work out what this distance is. So let’s call this distance 𝑥.

Now, in both cases, the object is initially moving. Therefore, it must have some kinetic energy. We can recall that the kinetic energy of an object 𝐸 sub 𝑘 is given by multiplying half by 𝑚 — the mass of the object — by 𝑣 — the velocity of the object — squared. So using this equation, we can work out in both cases the change in kinetic energy of the object as it slows to a stop. Let’s call the change in kinetic energy Δ 𝐸 sub 𝑘 because we use Δ to represent a change and 𝐸 sub 𝑘 represents our kinetic energy.

Now, in these two scenarios, the change in kinetic energy is not going to be the same. So let’s say for the first one our change in kinetic energy is Δ 𝐸 sub 𝑘 comma one. And for the second scenario, we’ve got Δ 𝐸 sub 𝑘 comma two. Now, the change in kinetic energy Δ 𝐸 sub 𝑘 in general for both cases is going to be the final kinetic energy 𝐸 sub 𝑘 final minus the initial kinetic energy Δ 𝐸 sub 𝑘 initial.

So for the first scenario Δ 𝐸 sub 𝑘 comma one, the final kinetic energy is half multiplied by the mass of the object multiplied by the final velocity of the object which is zero because the object is slowing to rest. And so this time becomes zero because we’ve got something multiplied by zero and that’s zero. So that’s the final kinetic energy.

And now, we can work out the initial kinetic energy. That’s simply becomes half multiplied by the mass of the object multiplied by the initial velocity squared. Now, the initial velocity is 𝑣. So we got half 𝑚𝑣 squared. And so the whole expression becomes zero minus half 𝑚𝑣 squared, which is simply negative one half 𝑚𝑣 squared. So that’s Δ 𝐸 sub 𝑘 comma one.

We can do the same for Δ 𝐸 sub 𝑘 comma two. Now, the final kinetic energy once again is going to be zero because even in this situation the object is coming to rest. And the final kinetic energy is going to be and the initial kinetic energy is going to be half multiplied by 𝑚 multiplied by the initial velocity. which in this case is five 𝑣 squared. Now, when we square five 𝑣, we get 25 𝑣 squared. Now here what we can do is to pull out the factor of 25. So we’re left with 25 times half 𝑚𝑣 squared. And the reason we’ve done that is because half 𝑚𝑣 squared kind of looks like this. And so overall, Δ 𝐸 sub 𝑘 comma two is equal to negative 25 times half 𝑚𝑣 squared.

If we wanted to, we could have also taken this negative sign in here so that the bit inside the parenthesis looks identical to Δ 𝐸 sub 𝑘 comma one. In fact, yeah, let’s do that. Now we’ve got Δ 𝐸 sub 𝑘 comma two is equal to 25 times negative half 𝑚𝑣 squared. This will come in handy later. For now though, let’s look at some other information.

We know that in both cases we’re using a force 𝐹 to decelerate the object. We can use this information to work out the amount of work done on the object in order to slow it down. We can recall that the amount of work done on an object 𝑤 is equal to the force applied to the object 𝐹 multiplied by the distance travelled by the object 𝑑. So we can calculate the amount of work done in the first scenario as 𝑤 is equal to 𝐹 the force multiplied by the distance 𝑠. And let’s call this 𝑤 sub one for the work done in the first scenario. 𝑤 sub two then becomes 𝐹 multiplied by the distance travelled which is 𝑥 and this 𝑥 is what we’re trying to find out here.

So why is this relevant? Well, we can use something known as the work energy theorem. The work energy theorem tells us that the amount of work done on an object is equal to the change in kinetic energy of that object. In other words, 𝑤 is equal to Δ 𝐸 sub 𝑘. So in each scenario, we can equate 𝑤 with Δ 𝐸 sub 𝑘.

So let’s do that for scenario one first. Negative one-half 𝑚𝑣 squared — that’s Δ 𝐸 sub 𝑘 comma one — is equal to 𝐹𝑠. That’s 𝑤 sub one. Now hang on, we’ve got a negative on one side, but a positive on the other? Well, no, we don’t. Remember the force is acting against the direction of motion. So the value of force is negative. And hence, we’ve got two negative values on each side of the equation.

Anyway, now let’s do the same thing for the second scenario. What we have here is 25 times negative one-half 𝑚𝑣 squared is equal to 𝐹𝑥. And at this point, we can see why it’s useful to have written the left-hand side as 25 times negative half 𝑚𝑣 squared. Because negative one-half 𝑚𝑣 squared is equal to 𝐹𝑠, so we can substitute that into here. So we have 25 times 𝐹𝑠 is equal to 𝐹𝑥. Let’s now divide both sides of the equation by 𝐹 so that all of these 𝐹s cancel out. And what we’re left with is 25𝑠 is equal to 𝑥.

And remember in this equation, we we’re trying to find out what the value of 𝑥 is. And now we’ve got it in terms of 𝑠, just the distance travelled in the first scenario. So we can calculate what 𝑥 is. 𝑥 is equal to 25 times 𝑠 which happens to be 12 meters which ends up being 300 meters.

And so we find out that our final answer is that the object moves 300 meters between the force first being applied and the object coming to rest.

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