Find speed given mass, starting speed and time

[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.