(III) A sled filled with sand slides without friction down a 3...

Question

(III) A sled filled with sand slides without friction down a 32° slope. Sand leaks out a hole in the sled at a rate of 2.0 kg/s. If the sled starts from rest with an initial total mass of 45.0 kg, how long does it take the sled to travel 120 m along the slope?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem, a stationary cart filled with grains moves down a frictionless ramp inclined at 20 degrees to the horizontal grains continuously fall out from a hole in the cart at a rate of 3.0 kg per second. Determine the time it takes for the cart to travel 90 m down the slope. If the total mass of the cart is 40 kg. Awesome. So that's our end goal is we're trying to figure out how long it will take for the cart to travel 90 m down this slope down this incline if the total mass of the cart is 40 kg. Awesome. So looking at our multiple choice answers, it appears that all of them are in the same units of seconds. So it looks like we're trying to find our time and units of seconds. So let's read them off to see what our final answer might be. A is 3.1 B is 4.4 C is 5.2 and D is 7.3. Awesome. So first off, let us consider that the positive X direction is along the incline in the downward direction as shown in our free body diagram. Free body diagram, which we can write or I should say which we can draw as. So I've gone ahead and drawn a free body diagram for this problem. As you can see, we have our cart on our incline and it's going down the slope. So it's going down the mountain, think of it that way. And as we could see in the going upwards from the center of our cart, we have our normal force and then pulling the cart back down to earth, keeping it on the slope, we have the weight force which is the mass of the cart multiplied by the acceleration due to gravity. And from our weight force vector, we can from the center, we have some angle theta and then we also have some angle theta for the slope or the incline, which we know it's 20 degrees to be the incline for the slope and it's traveling down in the horizontal direction. So the horizontal direction is X and the verti vertical direction is Y awesome. So now that we have a visualization of the forces at work here and have like a figure to help us visualize this problem. Let us call the position where the cart starts from the beginning of its motion to be X zero. And let's note that X zero is equal to 0 m. Let us also note that as grains leak out from the hole rather than are propelled out of the hole, there will be no relative velocity between the leaking grains and the cart during this process. And as a result, there will be no thrust force involved making this particular situation equivalent to one where the cart has no hope. So, using our free body diagram, we can see that the acceleration down the slope or the incline will be A X where A represents our acceleration is equal to G which represents the gravity. So G is gravity, but it's the acceleration due to gravity to be more exact multiplied by sine of drum roll theater. Because as we can see, our incline resembles a right triangle. Therefore, we can use trig identities to help us solve for this problem. So now we can recall and use the constant acceleration equations to help us solve for time. So let's call this equation one. And let's note that X equals X zero plus V zero X multiplied by T plus one half multiplied by AX multiplied by T squared. So A X multiplied by T squared. Fantastic. So now we need to note that X is the distance traveled down the slope or the incline. So we can add up here up above. We'll put a little set like a semi colon here. So X equals 90 m and we know that X zero equals 0 m. And we should also note that and let's write it off in blue off to the side here. So we don't get confused. We also know that the initial velocity V zero X is equal to 0 m per second because the cart starts from rest. So meters per second. Yeah. So the X components of the slow, so the X components of the initial velocity as it starts from rest. So that's what V zero X is, is the X component of the initial velocity since it starts from rest once again. So now we can write considering our acceleration that AX is equal to G. So now we can solve for acceleration. So A X equals G multiplied by sine of theta which is equal to 9.81 m per second squared multiplied by sine of 20 degrees which is equal to. So A X is equal to when we plug that into our calculator three point 3 5 5 2 m per second squared. Awesome. So with that in mind, we now need to solve for T and to solve for T, we need to rearrange equation one to isolate and solve for T, which is our final answer. So let me do that using a lot of algebra, we will get that T is equal to the square root of two multiplied by X divided by A X which is equal to when we plug in our known variables, two multiplied by 90 meters divided by our acceleration value, which we just found, which is 3.3552 m per second squared. And this is when we round to four decimal places. So we plug that into our calculator and round to one decimal place. We will get 7.3 seconds and that's our final answer. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter D 7.3 seconds. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

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(III) A sled filled with sand slides without friction down a 32° slope. Sand leaks out a hole in the sled at a rate of 2.0 kg/s. If the sled starts from rest with an initial total mass of 45.0 kg, how long does it take the sled to travel 120 m along the slope?

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