The 100 kg block in FIGURE EX7.24 takes 6.0 s to reach the floor after being released from rest. What is the mass of the block on the left? The pulley is massless and frictionless.

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Hey, everyone in this problem, we have two blocks connected with a cable through a pulley. The heavier block, 70 kg takes 5.5 seconds to cover 1.5 m from rest. We're asked to determine the mass of the lighter block and to assume that the pulley is frictionless and massless. We're given four answer choices. Option, a 68.6 kg. Option B 1.4 kg, option C 62.6 kg and option D 71.4 kg, we're gonna start by drawing a little diagram of what we have. So we have a pulley and connected to that police on a rope for cable. We have a block and we have the heavier block. We're gonna say the heavier block is on the left-hand side with a mass of 70 kg. And on the right, we have this smaller block or the lighter block and we wanna know what is its mass. We're gonna take up to be the positive direction and let's go ahead and draw free body diagrams for these blocks as well. So for the heavier block, give the force of gravity or the weight acting downwards So we're gonna call it F G H, the force of gravity for the heavier block. And we have the tension acting upwards from this cable and we're gonna call it T H the tension of the rope on that heavier block doing the same for the lighter block. We have the force of gravity for the lighter block acting downwards and the tension of the lighter block acting upwards. Now, we want to find the mass of this block. We know some information about these forces or we've drawn these forces out on our free body diagram. So let's recall Newton's second law. Newton's second law tells us that the sum of the forces is equal to the mass multiplied by the acceleration. So we know information with the some of the forces we're looking for the mass. So if we can find the acceleration of these blocks, we'll be able to use Newton's Second Law. Well, how do we find acceleration? Let's consider this as a kinematics problem. The initial speed of the block Is 0 m/s. And here we're talking about the heavier block and we're told this information about the heavier block. So the initial speed is zero m per second. We're told that it's released from rest. We don't know the final speed. The acceleration of the heavier block A H is what we're looking for. Now, we know this block travels 1.5 m. We need to figure out what direction that's going to happen in so that we can write the displacement. So the displacement, we're gonna call it delta Y it's going to be 1.5 m. We need to figure out the sign. Well, this block with mass 70 kg is heavier than the other block. So if we release it, it's going to fall, the force of gravity is going to be stronger because it's heavier. And so this block is going to fall, the block is falling or moving downwards, it's moving the opposite direction of what we've said. The positive direction is. And so we have a negative displacement, negative 1.5 m. And we're told that this takes 5.5 seconds. So we have three known values V nat delta Y and T we have one thing, we're trying to find the acceleration of this heavier block. We can go ahead and use our kinematic or U AM equations. We're gonna choose the equation that doesn't include the final velocity because we don't have information about that. And that's not what we're looking for. We get that delta Y is equal to VA T plus one half A H T squared. OK. So our displacement delta Y negative 1.5 m is equal to the initial velocity V knot multiplied by T, the initial velocity V knot is zero. So this term goes to zero and we have one half multiplied by the acceleration A H. And that's what we're looking for right now multiplied by the time squared. So we have 5.5 seconds squared. On the left hand side, negative 1.5 m. On the right hand side, we have one half multiplied by 5.5 seconds squared. This simplifies to A H multiplied by 15.125 second squared. And if we divide, we get the acceleration we were looking for which is negative 0. m per second squared, gonna draw a box around this value. So we don't lose track of it. So we found the acceleration of the heavier block. We found that it's negative, which makes sense because we've said this block is gonna be going downwards, it's gonna be accelerating downwards which is the opposite of our positive direction. Now, we can move over to Newton's law in the sum of the forces. We've drawn two free body diagrams, one for the heavier block and one for the lighter block. So we're gonna need to figure out two equations for each of these blocks. So we have the, the, some of the forces acting on the heavier block is equal to the mass of the heavier block multiplied by the acceleration of the heavier block. And we have that the sum of the forces acting on the lighter block is equal to the mass of the lighter block multiplied by the acceleration of the lighter block. Now the sum of the forces I recall our body diagrams, I'll go back up on our screen. So we can see we have tension acting upwards on both the heavier and the lighter blocks and we have the force of gravity acting downwards in the negative direction. And so when we write the sum of the horses in the, on the heavier block, we have the tension T H minus the force of gravity F G H. And this is equal to the mass of the heavier block multiplied by the acceleration of the heavier bulk. And we're gonna call this equation one. Um Actually, I'll simplify a little bit further. Let's write this in terms of the tension T H, we isolate the tension T H. We can write this is mh multiplied by A H plus F G H. And we're gonna call that equation one doing the same with our lighter block. We have the, the tension of the lighter block minus the force of gravity of the lighter block. OK. That's our sum of forces is equal to the mass of the lighter block multiplied by the acceleration of the lighter block. And remember we're looking for the mass of the lighter block M L. We're gonna call this equation two. Now, let's think about this problem. We have a pulley with a single cable that cable connects the heavier block and the lighter block. That means that the tension T H is going to equal the tension T L. OK? The tension is the same because it's the same cable, if the tension wasn't the same, that cable would snap or break. So we have T H is equal to T L. So now let's substitute equation one into equation two. Since T H is equal to T L, we can substitute our value that we found for T H as the value for T L. So we can write the mass mh multiplied by the acceleration A H plus the force of gravity F G H minus F G L is equal to the mass. M L multiplied by the acceleration A L. Now let's find out what these force of gravity are. Recall that the force of gravity is gonna be the mass multiplied by the acceleration due to gravity. So our equation becomes mh multiplied by A H plus mh multiplied by G minus M L multiplied by G is equal to M L multiplied by A L. No, again, we're looking for M L, we know the mass of the heavier block. We know the acceleration of the heavier block. We know the acceleration due to gravity. The value we don't have right now is the acceleration of the lighter B block. Well, this is very similar to the argument with attention. Hm The acceleration of the lighter block is gonna have the same magnitude of the accelera as the acceleration of the heavier block because they're connected by a cable. OK. The magnitude is gonna be the same, but the direction is gonna be opposite because when that heavier block falls, the lighter block is going to rise up. OK. So we have an acceleration constraint where the acceleration of the lighter block is equal to negative the acceleration of the heavier block. If we substitute this into our equation, we have the mass of the heavier block. Mh multiplied by A H plus mh multiplied by G minus M L multiplied by G is equal to M L multiplied by negative A H. Now we wanna isolate for M L. That's what we're looking for. So let's move this minus M L G term to the right hand side, then we get MH A H plus MH G is equal to M L G. And when we move it to the right hand side, it becomes positive minus M L multiplied by A H. And so solving for M L, we can factor, well, let's factor the mass on both sides. We have mh multiplied by A H plus G is equal to M L multiplied by G minus A H. Now we're gonna continue our problem on the right hand side here. And so M L, this value we're looking for the mass of the lighter block is going to be equal to the mass of the heavier block. Mh multiplied by the acceleration of the heavier block A H plus the gravitational acceleration G divided by gravitational acceleration G minus the acceleration of the heavier block. Now, let's substitute in the values. We know we have an expression for M L. We've done the hard part, we just need to substitute in our values. The mass of the heavier block is 70 kg. The acceleration of the heavier block is negative 0. 17 m per second squared. A plus a gravitational acceleration 9.8 m/s squared. And all of this is divided by gravitational acceleration, 9.8 m/s squared minus the acceleration of the heavier block, we have minus a negative value. So this is gonna give us plus 0.9917 m per second squared. And if you plug this all into your calculator, you're gonna get a mass of the lighter block of 68.6 kg. And that is the value we were looking for. That's what the question was asking us to find. One thing to note the mass of the heavier block was 70 kg. The mass of the lighter block we found was 68.6 kg. So it's not that much lighter. And that kind of makes sense. When we look at the acceleration, the acceleration was very small. And so if those blocks are close to the same weight, it makes sense that that acceleration is a small value. We compare to our answer choices. What we just found. We found that the mass of the lighter block in this system is 68.6 kg which corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

The 100 kg block in FIGURE EX7.24 takes 6.0 s to reach the floor after being released from rest. What is the mass of the block on the left? The pulley is massless and frictionless.

Verified Solution

Key Concepts

Newton's Second Law of Motion

Kinematics

Pulley Systems