Physics EK 1001: Question 362

[SKaminski]

Physics EK 1001: Question 362
....|
....|
..[4kg]

The block is lowered by a rope as shown. The tension T in the rope is 35 N. If the block is lowered 10 m, what is the change in kinetic energy of the block?

The credited answer is 50J.

Well, I assumed that (change)PE = (change)KE.

So I said mgh = (change)KE, which gave me 400 J. Obviously wrong.

The credited solution is: W = (change)KE , thus Fd = (change)KE.

To determine force we're taking T = 35, but mg = 40, so NET FORCE = 5N.
Again, Fd = (change)KE, so we have 5 N * 10 m = (change)KE = 50J.

And I have so many problems with this question on so many levels.

Why are we suddenly using net force? Why couldn't I use mgh here? I think it has something to do with conservative v non-conservative forces, but it seems like that rule is applied in odd situations with gravity, and I don't understand 100% yet.

---

Members don't see this ad.

 

[circulus vitios]

Notice that the tension on the string is less than the weight of the block. This means that there is a net force downward and that the block is accelerating downward. You can't use potential and kinetic energy equivalence because the block doesn't stop once it drops 10 meters.

 

[SKaminski]

Hey,

Thanks for taking the time to respond. My only question is:

Assuming that gravity is a conservative force in this scenario, can't we assume that this is an isolated system, and thus that energy is conserved?

 

[circulus vitios]

Hey,

Thanks for taking the time to respond. My only question is:

Assuming that gravity is a conservative force in this scenario, can't we assume that this is an isolated system, and thus that energy is conserved?

Sorry, I don't know why it doesn't work other than it doesn't work. Intuition told me that using kinetic and potential energy equivalence might lead to problems with an object that's still accelerating.

 

[cheesier]

The change would be 400J if the block were free falling, but the tension slows it down a bit. mgh is equal to work too, it's just the work done by gravity (F x d = mg x h). You would use this equation:

U1 - U2 + w(other) = K2 - K1

You are always implicitly using this equation, but in problems where the only force is gravity, you ignore the work(other) term. In this problem, work(other) = -35(10), so your equation simplifies to:

400 - 0 - 350 = K2 - K1

So you end up with K2 - K1 = 50