Find speed given mass, starting speed and time

[doublev231]

A cart which weighs 500g is moving on a horizontal table with the starting speed of 7 m/s. At the end of the cart there's a non-elastic wire attached, which is thrown over a pulley, and on the other end of the wire there's a 200g load. I need to find the speed of the cart and the distance it traveled after 5 seconds of movement

Given variables:
m1 = 500g
v (starting speed) = 7 m/s
m2 = 200g
t = 5s

I'm not sure what formula to use there. Any guidance would be appreciated.

 

[PeroK]

A cart which weighs 500g is moving on a horizontal table with the starting speed of 7 m/s. At the end of the cart there's a non-elastic wire attached, which is thrown over a pulley, and on the other end of the wire there's a 200g load. I need to find the speed of the cart and the distance it traveled after 5 seconds of movement

Given variables:
m1 = 500g
v (starting speed) = 7 m/s
m2 = 200g
t = 5s

I'm not sure what formula to use there. Any guidance would be appreciated.

You need to use the homework template and to do as much as you can yourself.

The first step is to draw a diagram. This should help you identify what is important in analysing the problem.

 

[doublev231]

You need to use the homework template and to do as much as you can yourself.

The first step is to draw a diagram. This should help you identify what is important in analysing the problem.

I tried drawing a diagram, but am still unsure what formula to use here. I need to calculate the speed of a cart moving in a horizontal line but I also need to account for the weight that's attached to the cart.

 

[PeroK]

I tried drawing a diagram, but am still unsure what formula to use here. I need to calculate the speed of a cart moving in a horizontal line but I also need to account for the weight that's attached to the cart.

That's true, but you still need to use the template.

By the way, my favourite formula is $F = ma$.

 

[doublev231]

That's true, but you still need to use the template.

By the way, my favourite formula is $F = ma$.

Okay, so I think I found one formula because you've mentioned the F=ma. Using it, we can get the velocity like this: v = Ft/m , and to get F we need m and a. We got m ( presumably I'll have to use the m1, which equals 500g ), and now we need a ( acceleration ). However, the problem is that the formula for acceleration is: a = v(final) - v(starting) / t and I don't have v final. How can I find the final v ( speed ) ?

 

[PeroK]

Okay, so I think I found one formula because you've mentioned the F=ma. Using it, we can get the velocity like this: v = Ft/m , and to get F we need m and a. We got m ( presumably I'll have to use the m1, which equals 500g ), and now we need a ( acceleration ). However, the problem is that the formula for acceleration is: a = v(final) - v(starting) / t and I don't have v final. How can I find the final v ( speed ) ?

So, your plan is:

1) Find the final speed; 2) Use this to find acceleration; 3) Use acceleration to find the force.

Can you see anything wrong with that?

What do you have in your diagram?

 

[doublev231]

So, your plan is:

1) Find the final speed; 2) Use this to find acceleration; 3) Use acceleration to find the force.

Can you see anything wrong with that?

What do you have in your diagram?

In my diagram I have drawn a cart that weighs 500 grams, and also marked that the starting speed of it is 7 meters per second. Then I have a wire that's over a pulley and at the end of the wire there's a weight that weighs in at 200 grams. Looks like I can't find the final speed without knowing the acceleration, is that what's wrong?

 

[PeroK]

In my diagram I have drawn a cart that weighs 500 grams, and also marked that the starting speed of it is 7 meters per second. Then I have a wire that's over a pulley and at the end of the wire there's a weight that weighs in at 200 grams. Looks like I can't find the final speed without knowing the acceleration, is that what's wrong?

Yes, so what's missing from your diagram?

 

[doublev231]

Yes, so what's missing from your diagram?

Looks like the acceleration of the cart is missing, however, I'm unsure how to calculate it, I need the final speed to do so. Currently I only have the time and the starting speed, and I'm unsure how to get the final one so I can calculate acceleration.

 

[PeroK]

Looks like the acceleration of the cart is missing, however, I'm unsure how to calculate it, I need the final speed to do so. Currently I only have the time and the starting speed, and I'm unsure how to get the final one so I can calculate acceleration.

It's more fundamental than that. What's the 200g mass doing in all this?

 

[doublev231]

It's more fundamental than that. What's the 200g mass doing in all this?

I think the 200 grams is slowing the cart down, thus reducing it's speed?

 

[PeroK]

I think the 200 grams is slowing the cart down, thus reducing it's speed?

I think you are misinterpreting the problem. As I see it, the 200g mass is hanging over a pulley, in mid air.

 

[doublev231]

I think you are misinterpreting the problem. As I see it, the 200g mass is hanging over a pulley, in mid air.

Yes, the 200g mass is hanging mid air over the pulley. So the 200g mass should add to the mass of the cart then, right?

 

[PeroK]

Yes, the 200g mass is hanging mid air over the pulley. So the 200g mass should add to the mass of the cart then, right?

What about gravity, then? Have you heard of that?!

 

[doublev231]

What about gravity, then? Have you heard of that?!

Oh, of course! So g = 9.81 m/s², is that what we need?

 

[doublev231]

What about gravity, then? Have you heard of that?!

Ooh, I get it now. We can now caculate F of the 200 g mass object like this: F = 200 * 9.81 = 1962 N, is that correct?

 

[PeroK]

Oh, of course! So g = 9.81 m/s², is that what we need?

Yes and no. Yes, gravity is exactly what you need. But, be careful, gravity is a force.

Let me point out something general about problems when you have two things joined together. If you have a trolley, say, and it's attached to a car and you try to pull the trolley. You could move the trolley on its own quite easily. But, if it's attached to the car, then in order to move the trolley, you have to move the car as well. Let's assume the trolley has a mass of $10kg$ and the car has a mass of $990kg$. And you apply a force of $10N$.

If you were pulling the trolley on its own, the acceleration would be $a_1 = 10N/10kg = 1ms^{-2}$

But, if the trolley is attached to the car, then the accleleration is $a_2 = 10N/1000kg = 0.01 ms^{-2}$

This is because the force must accelerate the whole system.

You always need to be careful, therefore, to identify everything that must move when a force is applied. It's a common mistake to calculate the acceleration based only on the mass of part of the system.

 

[PeroK]

Ooh, I get it now. We can now caculate F of the 200 g mass object like this: F = 200 * 9.81 = 1962 N, is that correct?

Yes, except the mass is only 200g, not 200kg!

Another general piece of advice is always to be careful with the units you are given.

 

[doublev231]

Yes, except the mass is only 200g, not 200kg!

Another general piece of advice is always to be careful with the units you are given.

Oh, good catch! It's actually 1.962 N. So now I know that the F of the object over pulley is 1.962 N. so now I can get the acceleration by adding the two weighs and then dividing the force by the sum of these objects? Like this: 1.962 N / 0.7 kg. If that's the case, would this mean that acceleration of the whole system is 2.8 m/s² ?

 

[PeroK]

Oh, good catch! It's actually 1.962 N. So now I know that the F of the object over pulley is 1.962 N. so now I can get the acceleration by adding the two weighs and then dividing the force by the sum of these objects? Like this: 1.962 N / 0.7 kg. If that's the case, would this mean that acceleration of the whole system is 2.8 m/s² ?

Yes, that looks good.

 

[doublev231]

Yes, that looks good.

And then I can use the formula for distance when given acceleration: s = v(starting) * t + 0.5 a * t². That way I will get the distance that the cart will travel in 5 seconds? Is that correct? And what about the speed after 5 seconds?

 

[PeroK]

And then I can use the formula for distance when given acceleration: s = v(starting) + 0.5 a * t². That way I will get the distance that the cart will travel in 5 seconds? Is that correct? And what about the speed after 5 seconds?

Once you have the acceleration, you have a choice of ways to find the distance. You don't actually need the final speed. The formula you suggest is good, although you have missed something out.

 

[doublev231]

Once you have the acceleration, you have a choice of ways to find the distance. You don't actually need the final speed. The formula you suggest is good, although you have missed something out.

I do need the final speed, the task asks to find the speed after 5 seconds and also the distance. I edited the formula, can you check it again? I added the v(starting) * t part. Is it complete now?

 

[PeroK]

I do need the final speed, the task asks to find the speed after 5 seconds and also the distance. I edited the formula, can you check it again? I added the v(starting) * t part. Is it complete now?

Yes, your formula is correct now.

Do you know another formula that would give you the final speed? I didn't notice it asked for that as well.

 

[doublev231]

I do need the final speed, the task asks to find the speed after 5 seconds and also the distance. I edited the formula, can you check it again? I added the v(starting) * t part. Is it complete now?

Once you have the acceleration, you have a choice of ways to find the distance. You don't actually need the final speed. The formula you suggest is good, although you have missed something out.

This is what I got right now: s = 7 * 5 + 0.5 * 2.8 * 25 = 70. I got 70 meters!

 

[doublev231]

Yes, your formula is correct now.

Do you know another formula that would give you the final speed? I didn't notice it asked for that as well.

is it a = v(final) - v(starting) / 2 ?

 

[PeroK]

is it a = v(final) - v(starting) / 2 ?

No, not divided by 2.

 

[doublev231]

That looks like average velocity to me.

I got 21 m/s. If I use v = u + at I also get 21 m/s. Isn't that the correct answer?

 

[PeroK]

I got 21 m/s. If I use v = u + at I also get 21 m/s. Isn't that the correct answer?

Yes. That all looks good. Well done!

 

[doublev231]

Yes. That all looks good. Well done!

Thank you very much for your help! I really appreciate it!

 

[doublev231]

Yes. That all looks good. Well done!

Hey, I got one more question. So the final results yielded this: distance = 70 meters and final speed = 21 meters per second. But the numbers seem off to me for some reason. The starting speed of the cart was 7 m/s and it had a weigh attached to it, and it traveled for 5 seconds. The current results suggest that the cart somehow gained speed! Is that really correct? I think I might have made a mistake somewhere.

 

[PeroK]

Hey, I got one more question. So the final results yielded this: distance = 70 meters and final speed = 21 meters per second. But the numbers seem off to me for some reason. The starting speed of the cart was 7 m/s and it had a weigh attached to it, and it traveled for 5 seconds. The current results suggest that the cart somehow gained speed! Is that really correct? I think I might have made a mistake somewhere.

Good spot. I think the question is probably that that the mass is slowing the cart down, rather than accelerating it. That's the problem with not seeing a diagram of the set-up. I assumed that the mass was pulling the cart in the direction of travel, but I guess the mass is supposed to be doing the opposite! That makes more sense, in fact.

Anyway, you can use the same technique to solve the correct problem.

 

[CWatters]

+1 the problem statement doesn't say which direction the cart is going at the start.

 

[doublev231]

Good spot. I think the question is probably that that the mass is slowing the cart down, rather than accelerating it. That's the problem with not seeing a diagram of the set-up. I assumed that the mass was pulling the cart in the direction of travel, but I guess the mass is supposed to be doing the opposite! That makes more sense, in fact.

Anyway, you can use the same technique to solve the correct problem.

Alright, so I suppose that the the force remains the same -> 1.962 N (F = ma = mg; F = 0.2 * 9.81;). So now I need to find the acceleration. Previously, we did this: a = 1.962 N / 0.7 kg. We added both masses and then divided by the power. But does the same rule apply here, now that the direction is changed?

 

[doublev231]

+1 the problem statement doesn't say which direction the cart is going at the start.

The problem stated that the cart is going to the left, but I'm unsure if that's useful. However, it said that the weight was attached to the end of the cart, so I presume the weight is supposed to slow the thing down.

 

[PeroK]

The problem stated that the cart is going to the left, but I'm unsure if that's useful. However, it said that the weight was attached to the end of the cart, so I presume the weight is supposed to slow the thing down.

Yes, it's just that the force now opposes the initial motion. So, $a = ?$

 

[doublev231]

Yes, it's just that the force now opposes the initial motion. So, $a = ?$

a = mass / force ? What about the distance formula? Will it remain the same? s = v(initial) * t + 0.5 * acceleration * time²

 

[PeroK]

a = mass / force ? What about the distance formula? Will it remain the same? s = v(initial) * t + 0.5 * acceleration * time²

All the formulas are the same. Note that these formulas involve Force, Acceleration, Velocity and Displacement and they are all vectors. For one-dimensional motion they can be positive or negative.

You can decide what direction is positive, but whatever direction is positive, the other must be negative.

 

[doublev231]

All the formulas are the same. Note that these formulas involve Force, Acceleration, Velocity and Displacement and they are all vectors. For one-dimensional motion they can be positive or negative.

You can decide what direction is positive, but whatever direction is positive, the other must be negative.

I'm somewhat stumped right now. So the formula for a is still FORCE / MASS , since you said all formulas remain the same. Is the force supposed to be negative now, since it opposes the thing now? But does that even make sense?

 

[PeroK]

I'm somewhat stumped right now. So the formula for a is still FORCE / MASS , since you said all formulas remain the same. Is the force supposed to be negative now, since it opposes the thing now? But does that even make sense?

A negative force very much makes sense. Just imagine throwing something up against gravity. If "up" is positive, then "down" is negative and gravity is a negative force that causes a negative acceleration.

 

[doublev231]

A negative force very much makes sense. Just imagine throwing something up against gravity. If "up" is positive, then "down" is negative and gravity is a negative force that causes a negative acceleration.

So does that mean the acceleration will be a = -force / mass, and all the calculations will remain the same except we'll get a negative value? But if that's the case, the distance will equal to 0, because s = 7 * 5 + 0.5 * (-2.8) * 5² = 35 - 35 = 0

 

[PeroK]

So does that mean the acceleration will be a = -force / mass, and all the calculations will remain the same except we'll get a negative value? But if that's the case, the distance will equal to 0, because s = 7 * 5 + 0.5 * (-2.8) * 5² = 35 - 35 = 0

Be careful. It is always $F = ma$. If the force negative, then so is the acceleration. On your diagram you need to clearly show the directions. I usually show the force with an arrow. So, if we assume to the right is positive, I would have:

$F \leftarrow$ (and such a force would be a negative number)

$F \rightarrow$ (and such a force would be a positive number)

You also need to be careful that $s$ is displacement (also a vector). So, $s =0$ tels you that the cart ended back where it started. The distance traveled is something different. Like if you go out for a walk and come back home, you've walked a certain distance even though you are back where you started.

 

[doublev231]

Be careful. It is always $F = ma$. If the force negative, then so is the acceleration. On your diagram you need to clearly show the directions. I usually show the force with an arrow. So, if we assume to the right is positive, I would have:

$F \leftarrow$ (and such a force would be a negative number)

$F \rightarrow$ (and such a force would be a positive number)

You also need to be careful that $s$ is displacement (also a vector). So, $s =0$ tels you that the cart ended back where it started. The distance traveled is something different. Like if you go out for a walk and come back home, you've walked a certain distance even though you are back where you started.

Oh, well, I suppose it kind of makes sense! The mass over the pulley constantly pulls the wire down due to gravity, and eventually the cart runs out of energy and returns back to it's original place because of the wire. So looks like I need to use a different formula to calculate the distance? Maybe s = 0.5at² then? Damn, I'm totally stumped :(

 

[PeroK]

Oh, well, I suppose it kind of makes sense! The mass over the pulley constantly pulls the wire down due to gravity, and eventually the cart runs out of energy and returns back to it's original place because of the wire. So looks like I need to use a different formula to calculate the distance? Maybe s = 0.5at² then? Damn, I'm totally stumped :(

In general, to find the distance you need to split the motion into the phases where it moves in each direction.

 

[doublev231]

In general, to find the distance you need to split the motion into the phases where it moves in each direction.

Okay, so if the cart returned back to its original place after 5 seconds, that means he moved forward and then he returned back. But I don't know what's the distance of these two directions that the cart completed. How do I figure that out? Also, talking about the speed. Since displacement is 0, does that mean that the final speed will also be 0?

 

[PeroK]

Okay, so if the cart returned back to its original place after 5 seconds, that means he moved forward and then he returned back. But I don't know what's the distance of these two directions that the cart completed. How do I figure that out? Also, talking about the speed. Since displacement is 0, does that mean that the final speed will also be 0?

Speed is the magnitude of velocity, so you can use the suvat equations to get the velocity and get the speed that way.

Distance is trickier. Can you think of a condition that represents a change of direction?

 

[doublev231]

Speed is the magnitude of velocity, so you can use the suvat equations to get the velocity and get the speed that way.

Distance is trickier. Can you think of a condition that represents a change of direction?

Hmm, perhaps if velocity equals 0, cart changes the direction ( because of the load over the pulley ) ? But how would that translate into a formula?

 

[PeroK]

Hmm, perhaps if velocity equals 0, cart changes the direction? But how would that translate into a formula?

The formula would be $v = 0$

 

[doublev231]

The formula would be $v = 0$

Wow, well, that was simpler than I thought! However, I'm still unsure how to get the total distance the cart traveled in those 5 seconds! The s formula you said is only used for displacement, what would be the formula for distance then?

 

[PeroK]

Wow, well, that was simpler than I thought! However, I'm still unsure how to get the total distance the cart traveled in those 5 seconds! The s formula you said is only used for displacement, what would be the formula for distance then?

Well, I'm going off line now. One way to look at this is that displacement is the "area" under a velocity against time graph. For displacement, the area above the x-axis counts as positive and the area below the x-axis is negative. To get the distance, you count all areas as positive.

You could draw a velocity/time graph for this motion.

Alternatively, you can find when the cart changes direction and calculate the displacement before and after that point. Then, to get the distance, you count all displacements as positive.