A 2.0-m-long, 500 g rope pulls a 10 kg block of ice across a hori...

Question

A 2.0-m-long, 500 g rope pulls a 10 kg block of ice across a horizontal, frictionless surface. The block accelerates at 2.0 m/s^2. How much force pulls forward on (b) the rope? Assume that the rope is perfectly horizontal.

Accepted Answer

Hey everyone in this problem, a metal chain 1.5 m long with a mass of kg is used to pull a 120 kg cart on a flat and smooth frictionless floor. The acceleration of the cart is 22 m/s squared. Were asked to determine the external force applied on the chain by the pulling motor. We're told to consider the chain to be horizontal. The answer choices were given are a Newtons. B 2640 Newtons C 2420 Newtons or D 220 Newtons. So let's just draw a little diagram of what we have. We have our cart here With a mass of 120 kg. And the chain With a length of 1.5 m is going to be used to pull that card and this train has its own mass of 10 kg and the chain is connected to a motor that's going to apply some extra. So let's go ahead and draw a free body diagram and let's start with the cart. Okay. We know the acceleration of the cart, we know the mass of the cart. So let's start with the free body diagram for the car. So for the cart, this is in contact with the ground. So we have the normal force pointing upwards, we have the force of gravity pointing downwards. We have the tension from that chain pointing to the left based on our diagram. Okay. That chain is pulling that cart, we have the tension from that chain and we're gonna take the left direction to be the positive direction. Now, we know that the sum of the forces in the X direction because that's where our motion is. It's going to be equal to the mass times the acceleration. That's Newton's second law. So some of the forces in the direction of motion, the direction the X direction, I'm gonna be equal to the mass times the acceleration, the forces we have acting in the extraction, the only force we have is attention due to that chain. And so the tension is going to be equal to the mass and we're talking about the cart. So 100 and 20 kg multiplied by its acceleration of 22 m per second squared, Which gives us a tension force t of 2640 mountains. All right, we have to be careful. Okay. We found a force. We found this tension for us. It can be easy to go back and answer the question and say this is the force we were looking for. But it's not, we're asked to find the external force applied on the chain by the pulling motor. This was the tension force applied from the chain on the cart. So let's go ahead and draw a free body diagram of the chain. Okay to see why these forces aren't the same. Okay. So if we draw a free body diagram of our chain, we have our chain and we know that we have that tension. In this case, it's acting to the right when we're looking at the chain that's pulling that crate or the car, and then we have the external force acting to the left and that external force. Yes, we have the tension pulling that cart, but the external force also has to be able to pull the weight of the chain because the chain has a mass. And so that external horse needs to be a little bit bigger in order to apply the tension force to the cart, but also be able to move the weight of that chain. Alright. So we're gonna apply Newton's second law. Again, we're gonna take the left direction to be positive just as we did with the cart and the sum of the forces in the X direction is going to be equal to the mass. And in this case, it's the mass of the chain, we're gonna call it M C and the acceleration of the chain. Now the sum of the forces is going to be the external force F external in the positive direction minus the tension T because now the tension T is pointing to the right, which is our negative extraction, this is equal to the mass of the chain Which is kg times its acceleration and it's going to have the exact same acceleration of that as that car. And so it's going to be 22 m/s squared. So our external force f external is going to be equal to the tension force. We move that to the right hand side, 2640 newtons plus 10 kg multiplied by 22 m/s squared. And recall that a Newton is equal to kilogram meter per second squared. So both of these terms have a unit of Newton, we add them together and we get 2860 newtons. And that is that external force applied on the chain by the pulling motor that we were looking for. If we go back up to our answer choices, we see that that corresponds with answer choice. A 2860 newtons. Thanks everyone for watching. I hope this video helped see you in the next one.

A 2.0-m-long, 500 g rope pulls a 10 kg block of ice across a horizontal, frictionless surface. The block accelerates at 2.0 m/s^2. How much force pulls forward on (b) the rope? Assume that the rope is perfectly horizontal.

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