If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Question 1a: 2015 AP Physics 1 free response

Free body diagram for two masses connected by string-pulley system.

## Want to join the conversation?

• At , why is the magnitude of the two tension forces the same? He says that it is because the tension is constant throughout the string due to the covalent bonds at the subatomic level. I find that hard to grasp. Could someone explain it a little bit further?
• Just know that since they are attached together by the string, the tension force will always be the same for both of them.
• Is it possible for the string to have different tension forces on different parts? For example, on Block 2, could the tension force be m1g since that is the gravitational force of Block 1, which is pulling on the string? Likewise, could the tension force on Block 1 be m2g since that is the gravitational force of Block 2, which is pulling the string?

This would make the tension forces different for each Block and using this logic works for part b of the FRQ too.
• A massless string has the same tension everywhere. We assume all strings are massless in high school physics, unless told otherwise.
(1 vote)
• At , why is the magnitude of the two tension forces the same? He says that it is because the tension is constant throughout the string due to the covalent bonds at the subatomic level. I find that hard to grasp. Could someone explain it a little bit further?
(1 vote)
• From my knowledge, we're assuming that the 2nd block will move downwards and that nothing will happen to the rope in the process. If the tensions of the string on both sides were different, the rope might snap or the blocks' accelerations might be different. I'm not entirely sure if this is the correct answer, and you might want to confirm with someone else.
(1 vote)
• At , why is the magnitude of the two tension forces the same? He says that it is because the tension is constant throughout the string due to the covalent bonds at the subatomic level. I find that hard to grasp. Could someone explain it a little bit further?
• Tension is the same everywhere in a massless string. If it weren't the same everywhere, then you would have one bit of string that has more force on one side than it does on the other, and since the string is massless, that bit of string would have infinite acceleration.

Since there is no such thing as a real massless string, it doesn't really make any sense to try to refer to covalent bonds as an explanation for why the tension is the same throughout the string!
• I noticed that he never mentioned the force of air resistance. Are we just assuming that it is negligible?
• Well actually it says in the question that m_2 = 2*m_1
Also... so the magnitude of tension is like arithmetic mean of the two masses?
Also... the tensions drawn in the answer are the same magnitude and direction, but actually they are opposite directions, because when you look at the string, since the pulleys are "negligible", you see that both point toward the center of the string, from different ends. So it makes sense to draw two tensions when you consider this as two systems, but you can also consider it one system, then there is no tension to take into account because it cancels out, and you have the overall acceleration equal m_1*g, or (m_2*g)/2, in the direction of the m_2 end of the string. But that doesn't answer the question since you have to draw two free body diagrams.
I'm just writing this so somebody could point out how wrong I am.
• Why did you say that tension is the same? Wouldn't it be the tension is different because there are different directions of acceleration? (That's what you said when you drew it on the blocks instead of the space provided for your answer)
• The tension is the same because it's the same string.
• The position of a bicycle is described by the following function:

x(t)=a−b⋅t+c⋅t2

with a=31 m, b=4 ms, c=1.1 ms2.

v(t1)= ? ,v(t2)= ?
Please show how you solve them in steps and what equation you use.. thank you
• To solve this you must be familiar with calculus. The rate of change of x is v hence differentiate the equation with respect to time.It should give you, -b +2c*t=v.Plug in the values of b and c and the time and voila! the answer.
(1 vote)

## Video transcript

- [Voiceover] Two blocks are connected by a string of negligible mass that passes over massless pulleys that turn with negligible friction, as shown in the figure above, we see that there. The mass m2 of block 2 is greater than the mass m1 of block 1. The blocks are released from rest. So let's just really quick think about what we think is going to happen here. Block 2 has a larger mass, they're connected by this string and pulley system here. Well if it has a larger mass, they're in the same gravitational field, it's going to have a larger weight and so the weight pulling down on block 2 is going to be larger than the weight pulling down on block 1 and so you're going to have block 2 accelerating downwards, you're going to have block 2 accelerating downwards and block 1 is going to accelerate upwards with the same magnitude. So block 1 is going to accelerate upwards and so with that just intuition of what we think is going to happen here let's try to tackle part A of this. So they tell us the dots below, these dots, represent the two blocks, so this represents block 1, this represents block 2. Draw free-body diagrams showing and labeling the forces, not components, exerted on each block. Draw the relative lengths of all vectors to reflect the relative magnitudes of all forces. Alright, so let's think about what are all the forces acting on each of these blocks. Well for each of these blocks you're going to have the force of gravity, you're going to have the weight of the blocks acting on them, so for example this first block m1, m1, and I'll draw it here first, what is the force of gravity going to be? The force of gravity is going to be its mass times the gravitational field. Now what is going to be the force of gravity on block 2? Well it's going to be larger than that so let me just draw it like this, so the force of gravity is going to be m2 times g, how did I know it was going to be larger? Well m2 has a larger mass and we are in the same gravitational field. Now we're not done there, no not done yet because now we also have the upward pulling force of the tension in the string and we could think about what the magnitude of that, we know that the, we know that the tension is going to be pulling upwards on both of the blocks but let's think about what its magnitude is going to be. In order for block 1 to accelerate upwards, which is in our intuition as to what would happen, the tension, the force of the tension has to be larger, the magnitude of the tension has to be larger than the magnitude of the weight. Now in order for block 2 to accelerate downwards the magnitude of the tension has to be less than the downward force of gravity, the upward force of tension has to be less than the downward force of gravity. So the magnitude of the tension is going to be inbetween the magnitudes of these two weights, so let me draw that. So the magnitude of the tension is going to be larger than m1g but smaller than m2g, so maybe like that, so that is tension right there and, whoops, and you're going to have on this side you're going to have the same magnitude of your tension pulling upwards but it's now less than the weight, which means that the block is going to accelerate downwards and we know that these two tensions, the magnitudes of these tension vectors are going to be the same because of the magnitude of the tension throughout this entire string is going to be the same. When you think about tension, this pulling force, you can think about what's happening at an atomic or a molecular level is these are the, you know the covalent bonds pulling on each other at an atomic level throughout the entire string and when we look at it on a macro level we view that as tension. So there you have it, I've drawn the forces on them but I haven't drawn them in the right place, I need to draw them on this, I need to draw them right over here. So let me draw that, I can actually draw a little bit more precisely because they gave us these lines. So on block 1 I have the weight so that is m1g, I also have the tension, I'll make that two of these lines tall, so I also have the tension. Now block 2 I have the same tension or same magnitude of tension I should say, now it's also pulling upwards, so tension, and I have its weight, which is going to be larger than the tension, so its weight, I'll make it three of these lines. The way I've drawn it m2 would have to be three times, m2 would have to be three times larger than m1 the way I've drawn it. They don't tell us that, they tell us that m2 is larger so this seems to be right. And we just want to get our intuition here since our tension, our upward force is larger than our downward force you're going to have a net upward force, which is going to accelerate block 1 up and here you're going to have a net downward force, which is going to accelerate block 2 down, which is in line with our intuition because block 2 weighs more than block 1.