Video: Calculating the Motion of Two Blocks Connected by a Rope

Two blocks are connected across a pulley by a rope as shown. The mass π‘šβ‚ of the block on the table is 4.0 kg and the mass π‘šβ‚‚ of the hanging block is 1.0 km. The mass of the rope is negligible. The table and the pulley are both frictionless. Find the acceleration of the system. Find the tension in the rope. Find the speed of the hanging block when it hits the floor. Assume that the hanging block is initially at rest and located 1.0 m vertically above the floor.

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Video Transcript

Two blocks are connected across a pulley by a rope as shown. The mass π‘š one of the block on the table is 4.0 kilograms. And the mass π‘š two of the hanging block is 1.0 kilogram. The mass of the rope is negligible. The table and the pulley are both frictionless. Find the acceleration of the system. Find the tension in the rope. Find the speed of the hanging block when it hits the floor. Assume that the hanging block is initially at rest and located 1.0 meters vertically above the floor.

From this last bit of information, we can add some detail to our diagram. We now know that π‘š two is a distance we can call 𝑑 of 1.0 meters above the floor. We’re told previously the masses π‘š one and π‘š two. And we want to solve overall for three pieces of information. We want to solve for the acceleration π‘Ž of the system, the tension 𝑇 in the cable connecting the two blocks, and the final speed, we can call 𝑣 sub 𝑓 of π‘š two, as it hits the ground. As we get started solving for system acceleration π‘Ž, we recall Newton’s second law of motion that the net force acting on an object or a system is equal to the mass of the object multiplied by its acceleration.

When we consider the forces involved with respect to our diagram, we see that the only force responsible for the motion of this system of masses is the gravitational force acting on π‘š two. That’s because our system overall is frictionless and the gravitational force on π‘š one is counteracted by the normal force pushing up on π‘š one. So by Newton’s second law, we can write that the weight force on π‘š two, which is π‘š two times 𝑔, equals the mass of our system, π‘š one plus π‘š two, multiplied by the acceleration of the system, π‘Ž. Rearranging this expression to solve for π‘Ž, we see it’s equal to π‘š two times 𝑔 divided by the sum of the masses π‘š one and π‘š two. 𝑔 we’ll treat as exactly 9.8 meters per second squared. And since we’re given π‘š one and π‘š two, we’re ready to plug in and solve for π‘Ž.

When we enter this expression on our calculator, we find that π‘Ž is 2.0 meters per second squared. That’s the overall acceleration of this system of masses. Next, we want to solve for the tension 𝑇 in the line connecting the two masses. The interesting thing about solving for 𝑇 is that since 𝑇, the tension, is a force on both masses, we could look at either one of the masses to solve for it. Just to pick a mass, let’s choose π‘š two. The two forces acting on π‘š two are the tension force 𝑇 and the weight force π‘š two times 𝑔. If we consider motion down to be motion in the positive direction, then Newton’s second law tells us we can write π‘š two times 𝑔 minus 𝑇 is equal to π‘š two times π‘Ž, where π‘Ž is the acceleration we solved for earlier.

Rearranging to solve for 𝑇, we find it’s equal to π‘š two times 𝑔 minus π‘Ž, or 1.0 kilograms multiplied by 9.8 minus 2.0 meters per second squared. When we multiply these values together, we find that 𝑇 is 7.8 newtons. That’s the tension in the line connecting the masses. And finally, knowing that π‘š two starts from rest and descends a distance 1.0 meters before it reaches the floor, we want to solve for 𝑣 sub 𝑓, its final speed, just as it does reach the floor. Since the mass π‘š two has a constant acceleration, we can use the kinematic equations to solve for its motion.

Specifically, we can use the equation that 𝑣 sub 𝑓 squared equals 𝑣 sub zero squared plus two times π‘Ž times 𝑑. Since π‘š two starts at rest, that means 𝑣 sub zero is zero. So our equation is simplified to 𝑣 sub 𝑓 squared equals two times π‘Ž times 𝑑. When we take the square root of both sides and plug in 2.0 meters per second squared for π‘Ž and 1.0 meters for 𝑑, when we calculate the square root, to two significant figures, it equals 2.0 meters per second. That’s the speed of block π‘š two just as it hits the ground.

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