An 8.00-kg block of ice, released from rest at the top of a 1.50-...

Question

An 8.00-kg block of ice, released from rest at the top of a 1.50-m-long frictionless ramp, slides downhill, reaching a speed of 2.50 m/s at the bottom. (a) What is the angle between the ramp and the horizontal?

Accepted Answer

Hey everyone in this problem, we have an experiment where we hold a stationary block of mass M equals 4.5 kg at the top of a 2. m long frictionless incline. We're going to release the block and measure its speed at the bottom of the ramp to be 4.1 m/s. They were asked to determine the angle between the horizontal and the surface of the incline. Alright, so let's try a little diagram. Okay, so we have this incline with an angle theta that we're trying to find, we're gonna have a block at the top, It has a mass of 4.5 kg And it's going to be stationary. The initial speed is zero and this incline All the way down to the end is going to be 2.3 m. I'm told that it's 2.3 m long incline. And then after some time we're gonna release the block, It's gonna come to the bottom of the ramp and at the bottom of the ramp it's gonna have the same mass 4.5 kg. And let me just, it's a little bit messy. 4.5 kg. And it's gonna have a speed, we'll call it V. F. The final speed of 4.1 m per second and then this is the same inclined the same angle theta. Alright, so what we wanna do is we want to find data so let's go ahead and draw a free body diagram of our problem first. So we have our block here. Okay we have the normal which we know points perpendicular to the surface and up we have our weight acting downwards the weight which is equal to M. G. Hey we have frictionless inclines, we don't have to worry about friction and then we have no other forces that are being acted on it. Okay? Alright so we're gonna take our frame of reference or our coordinate system to be like this, it's gonna be tilted. Okay so that it is parallel and perpendicular to our incline and when we do that we're going to have to decompose our weight into its Y. Component here and into its X. Component for our new coordinate system. Alright so we have our free body diagram. Now let's go ahead. We're trying to find the angle, we know that we should use the sum of forces. What we have. Is that the sum of the forces? Okay let's start in the X direction because that's the direction that we have motion is going to be equal to the mass times the acceleration in the extraction. Okay well what is this? Some of the forces in the extraction? In the extraction. We only have the weight in the X. Component of the weight. Okay. The X component of the weight. We're gonna call it W. X. And this is equal to M. A X. Now our call weight is MG. In order to get the X. Component. We do M G. Sine of theta and this is equal to M. A. X. And what you'll see is now we have this value of theta in our equation that we're trying to find. Okay, so that's great. Now we have a way to find that value. Okay, we have mass on both sides so we can divide by mass and we get G sign. Theta is equal to A X. Okay? And if we want to solve for theta can divide by G. Sign. Theta is equal to a X over G. Which tells us that theta is equal sign inverse of the acceleration in the extraction divided by G. Okay, the acceleration due to gravity. Alright, So we know G the acceleration due to gravity but we don't know the acceleration A. X. Hey, that's the acceleration in the direction that the block is moving and we know the velocity is. We don't know the acceleration. So let's recall. We can use our you am equations. Okay, we have the velocity initially Which is zero m/s. With the final velocity we're told is 4.1 m/s. Cave. This was included in our Diagram. We know that the distance the block travels is 2.3 m because we're told that that incline is 2.3 m long and we want to know the acceleration and we don't have information about time team. All right. So with this information we have three variables that we know. We have one. We don't want to know. Okay, so we're gonna one we do want to know. Sorry. A we want to know and t we don't have information about. So we're gonna choose the equation that doesn't include time. T. That's gonna be V f squared is equal to v naught squared plus two A. D. Okay, substituting our values 4.1 m per second. All squared is equal to zero m per second squared plus two a times 2.3 m. Okay, on the left hand side we get 16. m squared per second squared Is equal to on the right hand side. We get 4.6 m times the acceleration A. If we divide by 4.6 m we have meter squared per second squared. It's gonna leave us with a unit of meters per second squared, which is what we want for acceleration. We get that the acceleration a is equal to 3.654 m per second squared. And again that's the acceleration of the block. And because it's moving down the incline in the exact same direction that we have our X. Direction or the X. Component of our free body diagram. Okay, this is going to be equal to a X. So we can go back to find data using the acceleration we just found we get sine inverse of 3.654 meters per second squared divided by 9.8 m per second squared. And we get a theta value of 21.89°. And so that's the angle of the incline. If we go back up to our answer choices, Okay. We can approximate to one decimal place. We see that we have be the angle between the horizontal and the surface is 21.9°. Thanks everyone for watching. I hope this video helped see you in the next one.

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An 8.00-kg block of ice, released from rest at the top of a 1.50-m-long frictionless ramp, slides downhill, reaching a speed of 2.50 m/s at the bottom. (a) What is the angle between the ramp and the horizontal?

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