Question Video: Studying the Motion of a Body on a Rough Plane Connected through a Pulley to a Vertically Hanging Body at a Given Height | Nagwa Question Video: Studying the Motion of a Body on a Rough Plane Connected through a Pulley to a Vertically Hanging Body at a Given Height | Nagwa

Question Video: Studying the Motion of a Body on a Rough Plane Connected through a Pulley to a Vertically Hanging Body at a Given Height Mathematics • Third Year of Secondary School

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A body of mass 200 g rests on a rough horizontal table. It is connected by a light inextensible string passing over a smooth pulley, fixed to the edge of the table, to another body of the same mass hanging freely below the pulley 2 cm above the ground. The coefficient of friction between the table and the body resting on it is 1/3. Given that the system was released from rest, and the hanging body descended until it hit the ground, how much further did the body on the table travel until it came to rest? Take acceleration due to gravity 𝑔 = 9.8 m/sΒ².

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Video Transcript

A body of mass 200 grams rests on a rough horizontal table. It is connected by a light inextensible string passing over a smooth pulley fixed to the edge of a table to another body of the same mass hanging freely below the pulley two centimeters above the ground. The coefficient of friction between the table and the body resting on it is one-third. Given that the system was released from rest and the hanging body descended until it hit the ground, how much further did the body on the table travel until it came to rest? Take acceleration due to gravity 𝑔 to be equal to 9.8 meters per square second.

There is a lot of information in this question, so we will begin by drawing a diagram. We have two bodies of the same mass of 200 grams. The weight of these bodies is equal to the mass multiplied by gravity. As there are 1000 grams in one kilogram, these both have a downward force of 0.2𝑔. The tension throughout the string will be equal. And when the system is released, the freely hanging body will accelerate downwards and the body on the table will accelerate to the right. As the horizontal table is rough, there will be a frictional force acting against the body. There is also a reaction force going vertically upwards.

We know that the frictional force is equal to πœ‡ multiplied by the reaction force when πœ‡ is the coefficient of friction, in this case, one-third. Newton’s second law states that force is equal to mass multiplied by acceleration. We can use this to resolve vertically and horizontally. The body on the table is not moving vertically. Therefore, the forces up must equal the forces down. We have 𝑅 is equal to 0.2𝑔. The friction force must be equal to a third of this. This gives us a friction force of 49 over 75 newtons. Adding in the fact that the freely hanging body is two centimeters from the ground, we will now clear some space for the remainder of our calculations.

Resolving horizontally for the body on the table, we see that the sum of the forces is equal to 𝑇 minus 𝐹 r. This is equal to 0.2π‘Ž. We can substitute in the value of friction here. The freely hanging particle is accelerating downwards. This gives us the equation 0.2𝑔 minus 𝑇 is equal to 0.2π‘Ž. We have a pair of simultaneous equations that we can solve to calculate the acceleration π‘Ž. Adding equation one and equation two gives us 98 over 75 is equal to 0.4π‘Ž. Dividing both sides by 0.4 gives us an acceleration of 49 over 15. Once released, the system moves with an acceleration of 49 over 15 meters per second squared.

After the freely hanging body hits the ground, the string becomes slack and its tension is equal to zero. We can then calculate the acceleration of the body on the table after this point by solving negative 49 over 75 is equal to 0.2π‘Ž. Dividing both sides of this equation by 0.2 gives us π‘Ž is equal to negative 49 over 15. We notice that the acceleration after the string becomes slack is the negative acceleration prior to it.

Let’s now consider what is happening to the body on the table throughout its motion. We know that it starts from rest and accelerates uniformly at 49 over 15 meters per square second. During this time, it travels a distance of two centimeters as it must travel the same distance as body B. After the freely hanging body hits the ground, the body on the table decelerates. Its acceleration is equal to negative 49 over 15 meters per square second. We need to calculate how far it travels before it comes to rest. The final speed of the first part of the journey is equal to the initial speed of the second part.

We can now use our equations of uniform acceleration or SUVAT equations to calculate 𝑠. We know the following values for the two parts of the journey. We can now substitute these values into the equation 𝑣 squared is equal to 𝑒 squared plus two π‘Žπ‘ . In the first part, as 𝑒 is equal to zero, we have 𝑣 squared is equal to two multiplied by 49 over 15 multiplied by two. In the second part, as 𝑣 is equal to zero, we have zero is equal to 𝑣 squared plus two multiplied by negative 49 over 15 multiplied by 𝑠.

We can solve these simultaneously to eliminate 𝑣 squared. We can then divide our new equation by two and 49 over 15. Two minus 𝑠 is equal to zero, which means that 𝑠 is equal to two. The body travels another two centimeters until it comes to rest. This is the same distance as in the first part of the journey as the acceleration and then deceleration are equal.

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