Question Video: Determining Acceleration of a Bucket Using Pressure | Nagwa Question Video: Determining Acceleration of a Bucket Using Pressure | Nagwa

Question Video: Determining Acceleration of a Bucket Using Pressure Physics • Second Year of Secondary School

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In a well, a 0.35 m high bucket full to the brim with water of density 1000 kg/m³ is supported by a rope. The bucket is dropped into the well and its fall is slowed by the tension in the rope. During the bucket’s fall, the pressure on its base is 2310 Pa. At what rate does the bucket accelerate down the well?

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Video Transcript

In a well, a 0.35-meter-high bucket full to the brim with water of density 1000 kilograms per meter cubed is supported by a rope. The bucket is dropped into the well and its fall is slowed by the tension in the rope. During the bucket’s fall, the pressure on its base is 2310 pascals. At what rate does the bucket accelerate down the well?

To begin, it’s a good idea to draw a diagram to help show the information given to us in the question. We have a bucket that is 0.35 meters tall and full of water of density 1000 kilograms per cubic meter, which exerts a pressure of 2310 pascals on the bucket’s base. The bucket is attached to a rope and has been released down into a well. Because the bucket is attached to the rope, it falls more slowly than it would otherwise. It’s our job to find the bucket’s acceleration given this information.

Here, we have a situation in which a fluid is exerting a pressure on an object. In this case, the water in the bucket is exerting a pressure on the bucket’s base. Usually, we would think to use the equation 𝑃 equals 𝜌 times 𝑔 times ℎ, where 𝑃 is the pressure produced by the fluid, 𝜌 is the density of the fluid, 𝑔 is the acceleration due to gravity, and ℎ is the depth of the fluid.

However, in this case, we actually need to adjust this formula slightly. This is because the bucket, and therefore the water in the bucket, does not have an acceleration that is equal to 𝑔. There’s tension in the rope attached to the bucket, which slows the bucket’s fall and reduces its acceleration. So we need to replace this term 𝑔 with the bucket’s actual acceleration, which we’ll call 𝑎. This gives us 𝑃 equals 𝜌𝑎ℎ.

Since 𝑎 is the value we’re interested in solving for, we should rearrange this equation to make 𝑎 the subject. We can do this by dividing both sides of the equation by 𝜌ℎ. This way, the 𝜌ℎ terms on the right-hand side cancel out, leaving us with the expression 𝑎 equals 𝑃 divided by 𝜌ℎ.

Now, all we need to do is substitute in numeric values on the right-hand side. We know the pressure exerted on the base of the bucket, 𝑃, is equal to 2310 pascals. And the density of the water, 𝜌, is 1000 kilograms per cubic meter. We also know that the bucket stands 0.35 meters high and that it’s filled to the brim with water. This means that the depth of the water at the base of the bucket, ℎ, is also equal to 0.35 meters. Substituting these values in, we find that the acceleration 𝑎 is equal to 2310 pascals divided by 1000 kilograms per cubic meter times 0.35 meters.

Before we calculate though, let’s take a moment to check out the units here. We can recall that a pascal is equivalent to a kilogram per meter second squared. So, if we write it like this in the formula, it’s easier to see how the units will cancel each other out. Notice that kilograms cancel out of the numerator and denominator and that in the denominator we can cancel out a power of meters from that fraction. To simplify this, we can flip the fraction in the denominator and multiply it instead. So the units can be written as one over meters second squared times meters squared.

Now, another power of meters cancels out, leaving just meters per second squared. This is a good sign, since we are solving for an acceleration after all.

Finally, completing this calculation, we find that 𝑎 is equal to 6.6 meters per second squared. This is our final answer. We have found that when the bucket full of water is released down the well, it accelerates at a rate of 6.6 meters per second squared.

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