(II) Two masses are connected by a string as shown in Fig. 8–3...

Question

(II) Two masses are connected by a string as shown in Fig. 8–35. Mass m_A = 3.5 kg rests on a frictionless inclined plane, while m_B = 5.0 kg is initially held at a height of h = 0.75 m above the floor.

(c) Use conservation of energy to find the velocity of the masses just before m_B hits the floor. You should get the same answer as in part (b).

Accepted Answer

Welcome back. Everyone in this problem. A cargo sled with a mass of 4 kg made from a durable composite material is positioned on a frictionless aluminum incline at a 30 degree angle. It is connected through a smooth lightweight pulley to a vertical hanging steel crate that weighs 6 kg. The crate starts 1 m above the ground and we want to find the speed of the sled and create just before the crate hits the ground using energy conservation methods for our answer choices. A says it's 1.9 m per second. B 2.8 m per second. C 3.5 m per second and D 4.9 m per second. Now, what are we trying to figure out here? We're trying to find the speed of the sled off and the crate. Ok. So the speed of the entire system just before the crate hits the ground. So we're assuming that the system is isolated with no external non conservative forces doing work like no friction or air resistance. And we're assuming that the rope connecting the sled and create is in extensible and massless. Let's try to think about the types of energy involved. Now, initially, OK. Initially, we have the potential energy of our crated. OK. So that's what happens before the crate starts to move. So based on the principle of conservation of mechanical energy, OK, we know that the initial energy is going to be equal to the final energy. And so far, our initial energy is the potential energy of the crate based on gravity. OK. So let's call that UG crate or I think you can even write UGC. Now what happens after the crate falls? Well, no, OK. When the crate falls, then our sled would have moved by 1 m. That is just before it hits the ground. OK. So let me represent that distance here in our diagram. So a crate would have moved by about 1 m. OK? And now at this point, we're going to have the kinetic energy of the entire system just before our crate hits the ground. So let's just call that K plus the potential energy of our slit. Let's call that s OK. Now, in order to figure this out, we need to ask ourselves, what do we know about kinetic and potential energy? Well, we can recall recall that potential energy is equal to mass multiplied by gravity multiplied by height and kinetic energy is equal to half of the mass multiplied by the velocity squared. OK. So in that case, the potential energy of our crate UGC is going to be equal to the mass of the crate multiplied by gravity multiplied by the height that the crate was at. Ok. In this case, our height would be 1 m and that is going to be equal to our kinetic energy, which is a half of the mass of the system. OK. So that's the mass of the SLED plus the mass of the crate all multiplied by V squared plus the potential energy of the SLED. So that's the mass of the sled multiplied by gravity multiplied by the height of our sled. Ok. Now what do we know here? We know the mass of the crate 6 kg. We know that the height it was above just before it started to fall was 1 m. We also know the mass of the sled but we don't know the height that the SLED is at after it has moved a meter. Ok. I remember we're talking about the vertical height here because this is the height that compared to the force gravity directly to the force gravity. Now, how are we going to figure that out? Well, as we said, since it would have moved 1 m on our incline. Ok. And we want to know how much vertical height it moved by, then we can find the vertical height as a component of our 1 m distance. Ok. Nor incline recall that it was set at 30 degrees that 30 degrees is opposite to the vertical height. The height of the sled that we want. And now that means that vertical height, the height of the SLED is going to be equal to 1 m multiplied by the sine of 30 degrees. OK? The size of 30 degrees is a half. So the height, the height of the SLED, OK is really equal to a half of a meter. Now that we have that in mind, we can go ahead and solve for our speed V. So first let's rearrange our equation to solve for V and then substitute the variables. Now, if this is the case, let me come down in red here. OK, then that means our kinetic energy a half of MS plus MC multiplied by V squared is going to be equal to MC GH minus MS GHS. Since that's the case, we can multiply both sides by two. OK? And divide by MS MC. So that leaves us with V squared being equal to two multiplied by MCG minus MSG HS divided by MS plus MC. If we square root both sides, then that means that we, we've come over a bit to the left V is going to be equal to the square root of that expression. OK. So I can write that here. So that's two multiplied by MCG HK minus MS GHS. And all of that is divided by MS plus MC. Now we can go ahead and solve for our speed V and recall we're only taking a positive route because we're talking about the absolute value of the speed here. So if by substituting those values, that means we're going to have two multiplied by the mass of the crate 6 kg multiplied by the acceleration due to gravity. We will take that to be 9.8 m per second squared multiplied by the height that the crate had to fall, which is 1 m and we're going to subtract the mass of the sled, which is 4 kg multiplied by the acceleration due to gravity multiplied by the height, the vertical height that the SLED is low at compared to its original position, which is a half of a meter or 0.5 m. Ok? And then that means we're going to be subtracting all of or sorry, dividing all of that by the mass of the sled. Sorry, the mass of the cred 6 kg plus the mass of the SLED, which is 4 kg. So you know if we're going to solve, we can put this entire expression into our calculator. OK? I know when we do that, we should get that our speed. So two significant figures is 2.8 m per second. OK. Thus the speed, the speed of the sled and the crate just before the crate hits the ground is 2.8 m per second. If we look back at our answer choices, that means B is the correct answer. Thanks a lot for watching everyone. I hope this video helped

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