Elevator Tension Calculation

[polymerase]

omg I am so lost for this question,

look at this video first
http://mp.pearsoncmg.com/probhtml/applets/setup5_5_3.html

It shows an elevator with a small initial upward velocity being raised by a cable. The tension in the cable is constant. The energy bar graphs are marked in intervals of 600 J.

Question: Find the magnitude of the tension T in the cable. Be certain that the method you are using will be accurate to two significant figures.

 

[learningphysics]

Do they give the mass of the object?

From the applet. What is the initial velocity? What is the final velocity? What is the distance?

From these 3 you can get acceleration... then you can get the tension in the cable if you know the mass of the object, or the weight.

 

[polymerase]

Do they give the mass of the object?

From the applet. What is the initial velocity? What is the final velocity? What is the distance?

From these 3 you can get acceleration... then you can get the tension in the cable if you know the mass of the object, or the weight.

(v_2)^2 = (v_1)^2 + 2ad
0 = 16 + 2(a)(4)
a=2

U = mgh
2400 = m(9.8)(4)
m=61.22

f=ma
=2(61.22)
= 122.45 N

Is that right?

 

[learningphysics]

(v_2)^2 = (v_1)^2 + 2ad
0 = 16 + 2(a)(4)
a=2

a = -2

U = mgh
2400 = m(9.8)(4)
m=61.22

Good! I forgot about that 600J part... so you have the mass.

f=ma
=2(61.22)
= 122.45 N

Is that right?

Net force = ma

so T - mg = ma

from that you can get T... you need to use a = -2...

another way to do this problem (I didn't think of it before), and probably the better way is that you can say:

Work done by tension = final mechanical energy - initial mechanical energy

So
$$T*4 = mgh_{final} + (1/2)mv_{final}^2 - mgh_{initial} - (1/2)mv_{initial}^2$$

$$v_{final} = 0.$$

$$h_{initial} = 0$$

so this becomes

$$T*4 = mgh_{final} - (1/2)mv_{initial}^2$$

this way, you should get the same T as above... so it's a way you can double check your answer.

 

[polymerase]

a = -2



Good! I forgot about that 600J part... so you have the mass.



Net force = ma

so T - mg = ma

from that you can get T... you need to use a = -2...

another way to do this problem (I didn't think of it before), and probably the better way is that you can say:

Work done by tension = final mechanical energy - initial mechanical energy

So
$$T*4 = mgh_{final} + (1/2)mv_{final}^2 - mgh_{initial} - (1/2)mv_{initial}^2$$

$$v_{final} = 0.$$

$$h_{initial} = 0$$

so this becomes

$$T*4 = mgh_{final} - (1/2)mv_{initial}^2$$

this way, you should get the same T as above... so it's a way you can double check your answer.

THANK YOU THANK YOU THANK YOU...
just one aside question...the equation that u have on top there how come you multipled the tension by 4?
THANK YOU THANK YOU THANK YOU

 

[learningphysics]

no prob. the 4m is just the final height - initial height... 4 - 0 = 4m

it is the distance through which the tension acts... hence work by tension = T*4m

 

[polymerase]

no prob. the 4m is just the final height - initial height... 4 - 0 = 4m

it is the distance through which the tension acts... hence work by tension = T*4m

oh i see. ok so let me get this straight. is this statement true?

W = $$mgh_{final} + (1/2)mv_{final}^2 - mgh_{initial} - (1/2)mv_{initial}^2$$
but since W = F$$\Delta$$d, where F in this case is the tension, then therefore,
F$$\Delta$$d = $$mgh_{final} + (1/2)mv_{final}^2 - mgh_{initial} - (1/2)mv_{initial}^2$$

would you be able to help me on another physics problems??, its already posted, its called Impulse and average force

 

[learningphysics]

Yes, that's right. The way I keep track of it is... the forces that go in the left side of the equation are all the forces that aren't being dealt with as potential energy...

That's only T. although gravity is a force acting on the object, it is being dealt with as potential energy on the right side... so it doesn't go into the left side...

if I didn't want to use gravitational potential energy... I could say:

(T - mg)*4 = 1/2mvfinal^2 - 1/2mvinitial^2

this is mathematically exactly the same as before... but conceptually slightly different... I'm not choosing to deal with the gravitational force as energy...

it's always more convenient to use energies when you can.

of course change in kinetic energy must always be included on the right side.