EK 1001 physics # 267

[Oorham]

Hi everyone,

Could someone please explain the following question?
A 10 kg mass hangs by a rope. If the tension in the rope remains constant at 150 N, which of the following could be true?

A) The mass is at rest.
B) The mass has a constant velocity.
C) The mass is decelerating at 5 m/s2.
D) The mass is in dynamic equilibrium.

Apparently, C is the answer. In the book it says " the tension is greater than mg. The velocity of the mass must be changing in the upward direction. This leaves only answer C. If C is true, the mass is moving downward at any velocity and slowing down". I don't get this! I understand the first point, tension is higher, velocity is changing in the upward direction. If its going up, why is it saying that its moving downward??

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[AdaptPrep]

Hi, Oorham--

Signs in physics can make one go nuts at times.

The easiest way to solve this is draw a free-body diagram of the 10 kg mass. There is a force going upwards of 150 N (tension) and a force going downwards of 100 N (weight). Upward is positive sign and downward is negative sign. As you already understand, this means that the velocity has to be changing in the y-coordinate. Set up the equation:
F = ma
150 N - 100 N = (10 kg) (a)
50 N = 10a
5 m/s^2 = a

Since acceleration is positive 5 m/s^2, that means that the acceleration points upwards. This means one of two things: an object is accelerating upwards (increasing speed going upwards) or decelerating downwards (decreasing speed going downwards). They chose an example of the second thing.

It might help to remember:
final velocity - initial velocity = change in velocity
change in velocity / change in time = acceleration

If you are decreasing speed going downwards, then final velocity is less than initial velocity. Velocity in the downward direction is negative (-y).
Put in some numbers to make it more clear. Let's say an object is going downward at an initial velocity of -10 m/s. After one second, it has a final velocity of -5 m/s. The acceleration is:
[final velocity - initial velocity] / change in time = acceleration
[-5 m/s - (-10 m/s)] / 1 sec = +5 m/s^2

This can be called deceleration since the object is slowing down.

I hope that helps. Happy studying!

 

[Oorham]

Hi, Oorham--

Signs in physics can make one go nuts at times.

The easiest way to solve this is draw a free-body diagram of the 10 kg mass. There is a force going upwards of 150 N (tension) and a force going downwards of 100 N (weight). Upward is positive sign and downward is negative sign. As you already understand, this means that the velocity has to be changing in the y-coordinate. Set up the equation:
F = ma
150 N - 100 N = (10 kg) (a)
50 N = 10a
5 m/s^2 = a

Since acceleration is positive 5 m/s^2, that means that the acceleration points upwards. This means one of two things: an object is accelerating upwards (increasing speed going upwards) or decelerating downwards (decreasing speed going downwards). They chose an example of the second thing.

It might help to remember:
final velocity - initial velocity = change in velocity
change in velocity / change in time = acceleration

If you are decreasing speed going downwards, then final velocity is less than initial velocity. Velocity in the downward direction is negative (-y).
Put in some numbers to make it more clear. Let's say an object is going downward at an initial velocity of -10 m/s. After one second, it has a final velocity of -5 m/s. The acceleration is:
[final velocity - initial velocity] / change in time = acceleration
[-5 m/s - (-10 m/s)] / 1 sec = +5 m/s^2

This can be called deceleration since the object is slowing down.

I hope that helps. Happy studying!

Hello AdaptPrep, thank you very much for replying to my post. I guess what I don't understand is that how could an object move against the direction of net force? In this example, the net force points up, so does the acceleration. How could the object decelerate and go down?
hummm ok, as I'm typing now, sth came into my mind. Decelerate does not necessarily mean going down, right? The object could still be going up while its velocity decreases, i.e. decelerating! Am I right? I think my problem was with the terminology!

 

[AdaptPrep]

Hi, Oorham--

Deceleration is another way of saying your final velocity is less than your original velocity. Therefore, you are correct; that does not necessarily mean going down (it only means that your final velocity is less than your original velocity). You can decrease velocity while going up or while going down (or going right or going left). If something is falling, a net force upward will have to be employed to slow it down (decelerate it).

Also, to answer your question, when you shoot an arrow straight up, its velocity eventually decreases to zero (decelerates while going up because in this case, acceleration is directed downward of -9.8 m/s^2).

Let me know if that helps clarify it.

 

[Oorham]

Hi, Oorham--

Deceleration is another way of saying negative acceleration (your final velocity is less than your original velocity). Therefore, you are correct; that does not necessarily mean going down (it only means that your final velocity is less than your original velocity). You can decrease velocity while going up or while going down (or going right or going left). If something is falling, a net force upward will have to be employed to slow it down (decelerate it).

Also, to answer your question, when you shoot an arrow straight up, its velocity eventually decreases to zero (decelerates while going up).

A negative velocity means going down (or left) and positive velocity means going up (or right), but that does not apply to acceleration.

Let me know if that helps clarify it.

Yes! I got it! Thanks for taking the time

 

[shouldvestudied]

I don't see why answer A is not also be correct.

If the mass starts with velocity downward but accelerates upward at 5 m/s^2 then there will be a moment in time when the mass comes to rest (v = 0) before travelling upward.

 

[AdaptPrep]

Hi, shouldvestudied--

That is a good question. To be at rest, the mass cannot change position with time. In your scenario, its position will change in time since there is a net force acting on it. Newton's first law says an object will remain at rest if there is no net force on it. Since there is a net force on this mass, its position will change in time, and is, therefore, not at rest. (there is an instantaneous velocity of zero, but that does not remain constant) I hope that helps.