EK 1001 Physics #364

[dgbmonkey]

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If someone could please help me with this question, I would greatly appreciate it. (I have searched through the threads and did not see it posted already, so I apologize if I missed it somewhere).

The problem refers to a diagram of a 4-kg block held by a vertical rope with tension, T, and reads:

The block is lowered by a rope as shown. The tension T in the rope is 35 N. If the block begins at 5 m/s downward and is lowered 10 m, what is the approximate final velocity of the block?

A. 5 m/s
B. 7 m/s
C. 10 m/s
D. 12 m/s

The back of the book says the answer is B. 7 m/s, but I cannot figure out how they are getting this answer. This is how I am working the problem:

If T = 35 N and the force due to gravity = mg = 40 N, then the net force = 5 N in the downward direction. The work = the net force * the displacement = 5 * 10 = 50. I then set work equal to the change in kinetic energy, so 50 = 1/2mv^2 = 1/2m(vfinal-vinitial)^2. Solving for vfinal = sqrt(50*2/4) + vinitial = 5 + 5 = 10 m/s

What am I doing wrong here? 😕

 

[Postictal Raiden]

If someone could please help me with this question, I would greatly appreciate it. (I have searched through the threads and did not see it posted already, so I apologize if I missed it somewhere).

The problem refers to a diagram of a 4-kg block held by a vertical rope with tension, T, and reads:

The block is lowered by a rope as shown. The tension T in the rope is 35 N. If the block begins at 5 m/s downward and is lowered 10 m, what is the approximate final velocity of the block?

A. 5 m/s
B. 7 m/s
C. 10 m/s
D. 12 m/s

The back of the book says the answer is B. 7 m/s, but I cannot figure out how they are getting this answer. This is how I am working the problem:

If T = 35 N and the force due to gravity = mg = 40 N, then the net force = 5 N in the downward direction. The work = the net force * the displacement = 5 * 10 = 50. I then set work equal to the change in kinetic energy, so 50 = 1/2mv^2 = 1/2m(vfinal-vinitial)^2. Solving for vfinal = sqrt(50*2/4) + vinitial = 5 + 5 = 10 m/s

What am I doing wrong here? 😕

You were good up to unbolded part. You are complicating a simple problem.

Fnet = ma = 5 = 4a
so, a = 1.25m/s2

Now use this value to plug into this equation: V^2final = V^2initial + 2ax

V^final = 25 + 2(1.25m/s2)(10m) = 25 + 25 = 50

Vfinal = 7 m/s

 

[dgbmonkey]

You were good up to unbolded part. You are complicating a simple problem.

Fnet = ma = 5 = 4a
so, a = 1.25m/s2

Now use this value to plug into this equation: V^2final = V^2initial + 2ax

V^final = 25 + 2(1.25m/s2)(10m) = 25 + 25 = 50

Vfinal = 7 m/s

OH! That makes total sense. I hate when I make simple problems like these way more difficult. Thank you for your help.

I know that there are usually many different ways to work these problems, so why was my first method unsuccessful at obtaining the correct answer (even though I now know it wasn't the quickest method)? Would work not be equal to the change in kinetic energy? Or was there a problem with my values in the kinetic energy equation?

 

[dgbmonkey]

Nevermind, I think I figured it out. I believe the problem is that I did not consider the change in potential energy and its effect on work. Work would actually equal the change in KE + the change in PE.

W = 1/2m(vfinal-vinital)^2 + mg(hfinal-hinitial) = 1/2(4)(vfinal-5)^2 + (4)(10)(hfinal - (hfinal+10))
50 = 2(vfinal-5)^2 +40(-10)
10 = 2(vfinal-5)^2

Therefore, vfinal = 5 + sqrt(5), which is approximately equal to 7 m/s.

Regardless, obviously using the acceleration and method you suggested is much better, but I just wanted to make sure I was understanding the concepts in this question completely. Thanks again for your help.