# Question Video: Calculating Work, Velocity, and Acceleration Physics • 9th Grade

54 J of work is done when pushing an initially stationary wheeled desk of mass 48 kg across the floor of a hall. The desk is moved halfway across the 30 m length of the hall in a time of 20 seconds and then released. Thefriction of the wheels with the hall’s floor is negligible. How much work is done per meter that the desk is moved? What is the average acceleration of the deskover the distance that it moves? What is the velocity of the desk when it isreleased?

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### Video Transcript

54 joules of work is done when pushing an initially stationary wheeled desk of mass 48 kilograms across the floor of a hall. The desk is moved halfway across the 30-meter length of the hall in a time of 20 seconds and then released. The friction of the wheels with the hall’s floor is negligible. How much work is done per meter that the desk is moved?

Okay, so this question is about a wheeled desk being pushed across a floor. Let’s imagine that this here is the desk. And we’re told in the question that this desk is moved halfway across the 30-meter length of a hall. This means that the distance moved by the desk, which we’ve labeled as 𝑑, is equal to a half times 30 meters. This works out as a distance of 15 meters. Let’s add this distance to our diagram. We are told that the total amount of work done while pushing the desk over this distance is equal to 54 joules. Let’s label this work as 𝑊. This first part of the question is asking us to find how much work is done per meter that the desk is moved.

What we know is that 54 joules of work is done to move the desk a distance of 15 meters. Then the work done per meter must be equal to the total work done divided by the number of meters moved. Subbing in our values for the total work done and the number of meters moved, we get that the work done per meter is equal to 54 joules divided by 15. We should notice that the value in the denominator does not have units. That’s because this value is not the distance moved, but rather the number of meters moved. And the number of meters is simply 15. Evaluating this expression gives us a result of 3.6 joules. So our answer to this first part of the question is the amount of work done per meter that the desk is moved is 3.6 joules.

Okay, now let’s clear some space so that we can look at the second part of the question.

What is the average acceleration of the desk over the distance that it moves?

To answer this second part of the question and find the average acceleration of the desk, there are a couple of equations that it’s going to be helpful to recall. One of these two equations is Newton’s second law of motion, which says that the force 𝐹 acting on an object is equal to the object’s mass 𝑚 multiplied by its acceleration 𝑎. Since the question tells us that the desk has a mass of 48 kilograms, then this means that we know the value of the quantity 𝑚, so we can use this equation to find the acceleration of the desk if we know the force acting on it.

At the moment, we don’t know the value of this force. However, we can recall another equation that links the force to quantities that we do know the values of. Specifically, this equation says that 𝑊, the work done on an object by a force, is equal to the magnitude of the force 𝐹 multiplied by the distance 𝑑 moved by the object. In this case, we know the work done and the distance moved by the desk, so we can use those values in this equation to calculate the force 𝐹. Then we can sub that force along with the mass of the desk into this equation to calculate its acceleration.

We are told in the question that friction is negligible. This means that we can assume the force we’ll find from the work done using this equation is the only force acting on the desk. Okay, let’s make a start on these calculations. First off, we want to find the value of the force 𝐹. So let’s take this equation and rearrange it to make 𝐹 the subject. To do this, we divide both sides of the equation by the distance 𝑑. On the right-hand side, the 𝑑 in the numerator cancels with the 𝑑 in the denominator. This gives us an equation that says 𝑊 divided by 𝑑 is equal to 𝐹. Of course, we can also write this equation the other way around to say that the average force 𝐹 exerted on an object is equal to the work done 𝑊 divided by the distance moved by the object 𝑑.

We can now take our values for 𝑊 and 𝑑 and sub them into this equation. When we do this, we find that 𝐹 is equal to 54 joules divided by 15 meters. We might notice that this calculation looks remarkably similar to the one that we did in the first part of the question. The only difference is in the units because now the value in the denominator on the right-hand side has units of meters because it’s the distance moved by the desk, which is 15 meters. This difference in units means that our result isn’t going to have the same units of joules that we had in the first part of the question. Instead, since we have a work done in the SI unit of energy, joules, and a distance in the SI base unit of distance, meters, then this will give us a force 𝐹 with units of newtons, the SI unit of force.

Evaluating the expression for 𝐹 gives us a result of 3.6 newtons. So we now know the average force 𝐹 the acts on the desk and we also know the desk’s mass 𝑚. That means that we can use this equation from Newton’s second law of motion in order to calculate the average acceleration of the desk. But first, we need to rearrange the equation to make the acceleration 𝑎 the subject. To do this, we divide both sides of the equation by the mass 𝑚. On the right-hand side, the 𝑚 in the numerator cancels with the 𝑚 in the denominator. This gives us that 𝐹 divided by 𝑚 is equal to 𝑎, which we can also write the other way around to say that the acceleration 𝑎 is equal to the force 𝐹 divided by the mass 𝑚.

Subbing our values for 𝐹 and 𝑚 into this equation, we find that 𝑎 is equal to 3.6 newtons divided by 48 kilograms. We’ve got a force in the SI unit of newtons and a mass in the SI unit of kilograms. This means that the expression will give us an acceleration 𝑎 in the SI unit of meters per second squared. Evaluating the expression gives us a result of 0.075 meters per second squared. So our answer to the second part of the question is that the average acceleration of the desk over the distance that it moves is 0.075 meters per second squared.

Let’s now clear ourselves some space to look at the final part of the question.

What is the velocity of the desk when it is released?

Okay, so this last part of the question is asking us to find the desk’s velocity when it’s released after being pushed. In the previous part of the question, we found the average acceleration of the desk. The acceleration of an object is the rate of change of the object’s velocity over time. Mathematically, the average acceleration 𝑎 of an object is equal to the change in the object’s velocity Δ𝑣 divided by the change in time Δ𝑡 over which the velocity change occurs. We are told in the main text of the question that the object is pushed for a time of 20 seconds. This is the value for Δ𝑡, the time interval over which the desk’s velocity is changing.

We are also told in the question that the desk is initially stationary. The change in the velocity of the desk Δ𝑣 must be equal to its final velocity 𝑣 subscript 𝑓 minus its initial velocity 𝑣 subscript 𝑖. Since we know that the desk starts out stationary, this means that its initial velocity has a value of zero meters per second. Okay, if we take this equation for the acceleration, we can multiply both sides of it by Δ𝑡. When we do that, we get this equation here. And on the right-hand side, the Δ𝑡 in the numerator cancels the Δ𝑡 in the denominator.

Then we can also write this equation the other way around to say that the change in an object’s velocity Δ𝑣 is equal to the object’s average acceleration 𝑎 multiplied by the time interval Δ𝑡. If we sub the desk’s average acceleration and the time interval over which it accelerates into this equation, we have that the desk’s velocity changes by 0.075 meters per second squared multiplied by 20 seconds. This works out as 1.5 meters per second. We know that the desk’s velocity changes by 1.5 meters per second and that its initial velocity is zero meters per second.

If we take this equation for Δ𝑣 and add the initial velocity 𝑣 subscript 𝑖 to both sides, then on the right-hand side, the plus and minus 𝑣 subscript 𝑖’s cancel each other out. This gives us an equation that says that the desk’s final velocity 𝑣 subscript 𝑓 is equal to its initial velocity 𝑣 subscript 𝑖 plus the change in velocity Δ𝑣. Since in this case we know that the initial velocity is zero meters per second, then the final velocity of the desk is simply equal to its change in velocity. This value is 1.5 meters per second. So our answer to the final part of the question is that the velocity of the desk when it’s released is 1.5 meters per second.