Find speed given mass, starting speed and time

[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.

 

[doublev231]

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.

Alright, I'll see what I can come up with. Thank you very much for sacrificing your time to help me out! You've helped me out a ton. Have a nice day!

 

[doublev231]

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.

This is what I got so far:

F = 1.962 N
a = -2.8 m/s² ( because it's de-accelerating )
t ( to stop ) = 7 m/s / 2.8 m/s² = 2.5s
s = 7m * 2.5s * 2 ( because it will go forward and back) = 35 meters
v(final) = 0 m/s

does this look correct to you?

 

[PeroK]

This is what I got so far:

F = 1.962 N
a = -2.8 m/s² ( because it's de-accelerating )
t ( to stop ) = 7 m/s / 2.8 m/s² = 2.5s
s = 7m * 2.5s * 2 ( because it will go forward and back) = 35 meters
v(final) = 0 m/s

does this look correct to you?

Not quite. It takes $2.5s$ to stop. So you need to work out the distance covered in that time. Note that because motion is all in one direction, displacement equals distance.

Then, you have to look at the remaining $2.5s$.

Another thing to learn is how to use the suvat equations when you break a problem up into different phases. I would do something like this:

Let $s_1$ be the displacement for the first $2.5s$ and $s_2$ be the displacement during the second $2.5s$.

Note that when you do this, the final velocity for the first phase becomes the initial velocity for the second phase. In this case, that velocity is zero.

 

[doublev231]

Not quite. It takes $2.5s$ to stop. So you need to work out the distance covered in that time. Note that because motion is all in one direction, displacement equals distance.

Then, you have to look at the remaining $2.5s$.

Another thing to learn is how to use the suvat equations when you break a problem up into different phases. I would do something like this:

Let $s_1$ be the displacement for the first $2.5s$ and $s_2$ be the displacement during the second $2.5s$.

Note that when you do this, the final velocity for the first phase becomes the initial velocity for the second phase. In this case, that velocity is zero.

Okay, this is what I tried to get the distance traveled in 2.5 seconds:
t(to stop) = 7 / 2.8 = 2.5s
s = 7m/s * 2.5s + 0.5 * (-2.8m/s²) * 2.5s² = 17.5 - 8.75 = 8.75 meters

and to calculate s2 I will do the same except this time the acceleration will be 2.8 m/s² and initial speed = 0, correct? Then add those two together to get total distance.

Does this look ok to you?

 

[PeroK]

Okay, this is what I tried to get the distance traveled in 2.5 seconds:
t(to stop) = 7 / 2.8 = 2.5s
s = 7m/s * 2.5s + 0.5 * (-2.8m/s²) * 2.5s² = 17.5 - 8.75 = 8.75 meters

and to calculate s2 I will do the same except this time the acceleration will be 2.8 m/s² and initial speed = 0, correct? Then add those two together to get total distance.

Does this look ok to you?

Yes.

 

[doublev231]

Yes.

s1 equals = 8.75 meters, s2 equals = 8.75 meters. In total I got 17.5 meters! Yay! Thank you so much!

So, another task was final speed. Does this mean that the final speed will be equal to 0 meters per second?

 

[PeroK]

s1 equals = 8.75 meters, s2 equals = 8.75 meters. In total I got 17.5 meters! Yay! Thank you so much!

So, another task was final speed. Does this mean that the final speed will be equal to 0 meters per second?

You have to a) think more clearly and carefully about the motion and b) use the formulas.

Note: $s_2 = - 8.75m$, as it is a displacement. The distance $d_2$, say, is $8.75m$.

Overall, $s = s_1 + s_2 = 0m$ and $d = d_1 + d_2 = 17.5m$

 

[doublev231]

You have to a) think more clearly and carefully about the motion and b) use the formulas.

Note: $s_2 = - 8.75m$, as it is a displacement. The distance $d_2$, say, is $8.75m$.

Overall, $s = s_1 + s_2 = 0m$ and $d = d_1 + d_2 = 17.5m$

Oh, okay, I think this is the formula, right? v = u + at
Therefore, v = 0 m/s + 2.8 m/s² * 2.5s = 7 m/s
Is that correct? The cart returned back to it's original speed?
But does that make sense, because after 5 seconds, it's supposed to stop, so how can it have a velocity anymore?

 

[PeroK]

Oh, okay, I think this is the formula, right? v = u + at
Therefore, v = 0 m/s + 2.8 m/s² * 2.5s = 7 m/s
Is that correct? The cart returned back to it's original speed?

Yes, but note that you have changed your sign convention. Now, you have motion to the left as positive. That's all right as long as you make it clear what you are doing.

It was probably better to calculate the final velocity, which, in fact, you could do directly by:

$v_f = v_i + at = 7m/s + (-2.8m/s^2)(5s) = -7m/s$

And that also gives you the final speed of $7m/s$.

From this you can see that the relationship between velocity and speed is simple. It's the relationship between displacement and distance that you have to be more careful with.

 

[doublev231]

Yes, but note that you have changed your sign convention. Now, you have motion to the left as positive. That's all right as long as you make it clear what you are doing.

It was probably better to calculate the final velocity, which, in fact, you could do directly by:

$v_f = v_i + at = 7m/s + (-2.8m/s^2)(5s) = -7m/s$

And that also gives you the final speed of $7m/s$.

From this you can see that the relationship between velocity and speed is simple. It's the relationship between displacement and distance that you have to be more careful with.

Can I also write it like this: ... = |-7 m/s| = 7 m/s ?
Also, if after 5 seconds it completely stops, then how can it have a velocity anymore? Shouldn't it be 0?

 

[PeroK]

Can I also write it like this: ... = |-7 m/s| = 7 m/s ?
Also, if after 5 seconds it completely stops, then how can it have a velocity anymore? Shouldn't it be 0?

It stops after $2.5s$ then it gets pulled backwards. That's what the equations are telling you. But, also, that's what you should be able to see from an understanding of the motion.

 

[PeroK]

Can I also write it like this: ... = |-7 m/s| = 7 m/s ?

If you want to be pedantic then you can show that you got the speed by taking the absolute value of the velocity. But, I'd just write down the positive number as the speed. I think that is a simple enough step.

 

[doublev231]

If you want to be pedantic then you can show that you got the speed by taking the absolute value of the velocity. But, I'd just write down the positive number as the speed. I think that is a simple enough step.

Looks like that's it then! Thank you very much once again for going through all of this with me! You've done me a huge favor. Thank you a ton!

 

[YarnMonkey]

In this diagram, we see that the .5kg cart is being pulled by the .2 kg mass and has an initial velocity of 7m/s.
We know that the net force (ΣF) acting on the system is going to be the .2 mass being pulled down by the force of gravity, as there is no friction between the cart and the surface. We also know that acceleration is conserved within a system, or that the .2kg mass and .5kg cart accelerate at the same rate, despite the initial velocity.
This can be written in formula as:
ΣF = Fg (.2kg mass) = (.5+.2)*a = (.2*9.8)
a = (.2*9.8)/(.2+.5) = 2.8 m/s/s
I replaced the ΣF with ma because F = m*a and I used both masses because acceleration is conserved within the system meaning that in order to calculate the acceleration of the entire system I need to add the masses to count for all masses within the system. In other words, my formula describes all forces acting on every mass within a system, so in order to calculate the acceleration I need to replace the mass in the net force (ΣF) with all masses because I wrote an equation that applies to all masses in the system.

Now that we know the acceleration of both masses, we can use kinematics to solve for the distance traveled. This can be done with the equation:
Δx = (vi*t)+½a(t*t)
Δx = (7*5)+½2.8(5*5) = 70m

Now that we know the distance, we can find the speed using this other equation:
(vf*vf)=(vi*vi)+2aΔx
vf = √(vi*vi)+2aΔx
vf = √(7*7)+2(2.8)(70) = 21 m/s

I hope that helps!

 

[haruspex]

View attachment 217405
In this diagram, we see that the .5kg cart is being pulled by the .2 kg mass and has an initial velocity of 7m/s.
We know that the net force (ΣF) acting on the system is going to be the .2 mass being pulled down by the force of gravity, as there is no friction between the cart and the surface. We also know that acceleration is conserved within a system, or that the .2kg mass and .5kg cart accelerate at the same rate, despite the initial velocity.
This can be written in formula as:
ΣF = Fg (.2kg mass) = (.5+.2)*a = (.2*9.8)
a = (.2*9.8)/(.2+.5) = 2.8 m/s/s
I replaced the ΣF with ma because F = m*a and I used both masses because acceleration is conserved within the system meaning that in order to calculate the acceleration of the entire system I need to add the masses to count for all masses within the system. In other words, my formula describes all forces acting on every mass within a system, so in order to calculate the acceleration I need to replace the mass in the net force (ΣF) with all masses because I wrote an equation that applies to all masses in the system.

Now that we know the acceleration of both masses, we can use kinematics to solve for the distance traveled. This can be done with the equation:
Δx = (vi*t)+½a(t*t)
Δx = (7*5)+½2.8(5*5) = 70m

Now that we know the distance, we can find the speed using this other equation:
(vf*vf)=(vi*vi)+2aΔx
vf = √(vi*vi)+2aΔx
vf = √(7*7)+2(2.8)(70) = 21 m/s

I hope that helps!

If you read through the thread you will see that doublev231 came up with that solution earlier, then realized that the initial speed is supposed to be away from the pulley.