Question Video: Studying the Motion of Two Bodies Hanging Freely Connected by a String Passing through a Pulley Mathematics

Two masses π‘š and 88 g are attached to the ends of a light string passing over a smooth pulley. Determine the value of π‘š, given that when the system was released, the other mass descended 11.76 m in 2 seconds. Take the acceleration due to gravity 𝑔 = 9.8 m/sΒ².

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Video Transcript

Two masses π‘š and 88 grams are attached to the ends of a light string passing over a smooth pulley. Determine the value of π‘š, given that when the system was released, the other mass descended 11.76 meters in two seconds. Take the acceleration due to gravity 𝑔 equal to 9.8 meters per square second.

We begin by modeling the system as shown. We are told that the masses are equal to π‘š and 88 grams. We recall that there are 1,000 grams in one kilogram, and as the gravity is given in meters per square second, we need the mass to be in kilograms. 88 grams is equal to 0.088 kilograms. This means that the downward forces of the two bodies are π‘šπ‘” and 0.088𝑔. We have a light string and a smooth pulley, which means that the vertical tensions will be equal. The entire system will move with the same acceleration, and we are told that the 88-gram body descends. This means that the body of mass π‘š will accelerate upwards.

We can now use Newton’s second law, 𝐹 equals π‘šπ‘Ž, to create equations for body A and body B. Body A is accelerating vertically upwards. Therefore, the sum of its forces is equal to 𝑇 minus π‘šπ‘”. This is equal to π‘šπ‘Ž. Body B is accelerating downwards. Therefore, the sum of its forces is 0.088𝑔 minus 𝑇. We assume the downward direction is positive. This is once again equal to the mass multiplied by the acceleration. At this stage, we have three unknowns: the tension, the mass of body A, and the acceleration of the system. We will need to use the information from the question that tells us that the body of 88 grams descends 11.76 meters in two seconds.

We can use the equations of uniform acceleration to calculate the acceleration. These are often referred to as the SUVAT equations, where 𝑠 is the displacement β€” in this case equal to 11.76 meters. 𝑒 is the initial velocity, 𝑣 is the final velocity, π‘Ž is the acceleration, and 𝑑 is the time, in this question, two seconds. We know that 𝑠 is equal to 𝑒𝑑 plus a half π‘Žπ‘‘ squared. Substituting in our values, we have 11.76 is equal to zero multiplied by two plus a half multiplied by π‘Ž multiplied by two squared. The right-hand side simplifies to two π‘Ž, so this is equal to 11.76. Dividing both sides of this equation by two gives us π‘Ž is equal to 5.88. The acceleration of the system, once it is released, is 5.88 meters per square second.

Substituting this as well as 𝑔 equals 9.8 into our equation for body A gives us 𝑑 minus 9.8π‘š is equal to 5.88π‘š. Adding 9.8π‘š to both sides of this equation gives us 𝑑 is equal to 15.68π‘š. We will call this equation one. Substituting the value for the acceleration into the equation for body B gives us 0.8624 minus 𝑇 is equal to 0.51744. Rearranging this equation gives us a value of 𝑇 equal to 0.34496. We can now substitute the tension, which is equal to 0.34496 newtons, into equation one. This tension force is equal to 15.68π‘š. Dividing both sides by 15.68 gives us π‘š is equal to 0.022. This is the mass of the body in kilograms. By multiplying this answer by 1,000, we can convert to grams, which is the same units as the other mass. The mass π‘š is equal to 22 grams.

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