Tension Force Problem (applying Work-Kinetic Energy Theorem)

• Heisenberg7
Heisenberg7
Homework Statement
Figure 7-41 shows a cord attached to a cart that can slide along a frictionless horizontal rail aligned along an x axis. The left end of the cord is pulled over a pulley, of negligible mass and friction and at cord height h = 1.20 m, so the cart slides from x_1 = 3.00 m to x_2 =1.00 m. During the move, the tension in the cord is a constant 25.0 N. What is the change in the kinetic energy of the cart during the move? (Source: Fundamentals of Physics, Halliday and Resnick)
Relevant Equations
##\Delta K = W##

We know that net work done is equal to the change in kinetic energy, so we write: $$\Delta K = W$$ The tension is acting at angle ##\theta## due to the x axis, so we will only be taking its x component ##T_x = T \cos{\theta}##. Since we can look at this as one dimensional motion (##T_y## does no work, only ##T_x## does; only tension does work), we can write: $$W = \int_{x_1}^{x_2} T_x \, dx \implies W = \int_{x_1}^{x_2} T \cos{\theta} \, dx \implies W = T \int_{x_1}^{x_2} \cos{\theta} \, dx$$ How does one go about solving this integral? I've seen someone's solution and it says that the work done by tension is ##W = Td## where ##d = \sqrt{x_1^2 + h^2} - \sqrt{x_2^2 + h^2}##. So, I guess my integral should be equal to this, but I have no idea how to solve it.

DeBangis21
You can do it with the integral, but it's easier if you think about the tension ##T## on the left side of the pulley and how far its displaced as the mass goes from ##x_1## to ##x_2##.

Are you seeing that (it's how the solution you saw is derived), or are you explicitly interested in evaluating the integral?

Lnewqban
erobz said:
You can do it with the integral, but it's easier if you think about the tension ##T## on the left side of the pulley and how far its displaced as the mass goes from ##x_1## to ##x_2##.
Oh, hmm, I'm not sure if I can see it. Could you elaborate? Also, did I setup my limits of integration correctly?

Heisenberg7 said:
Oh, hmm, I'm not sure if I can see it. Could you elaborate? Also, did I setup my limits of integration correctly?
To do the integral you must change variables. If you want to go over the math/calculus, we should do that after you see where the solution you saw comes from.

What is the tension in the rope on the left side of the pulley?

What is the angle between ##T## and its displacement over there?

Draw lines from ##x_1## to the top of the pulley, repeat for ##x_2##

How far has ##T##(on the left side) been displaced using this information (and an inextensible rope approximation)?

Last edited:
erobz said:
To do the integral you must change variables. If you want to go over the math/calculus, we should do that after you see what the solution you saw comes from.

What is the tension in the rope on the left side of the pulley?

What is the angle between ##T## and its displacement over there?

Draw lines from ##x_1## to the top of the pulley, repeat for ##x_2##

How far has ##T##(on the left side) been displaced using this information (and an inextensible rope approximation)?
Oh, I see it now. So on the left side of the pulley, the tension is going to be the same, so ##T##. The angle between ##T## and its displacement is going to be 0°. Well, it will have travelled "the length of the rope at ##x_1##(from the top of the pulley to ##x_1##)" - "the length of the rope at ##x_2##(from the top of the pulley to ##x_2##)" which is exactly what ##d## is.

erobz
Could we do the math now?

Heisenberg7 said:
Could we do the math now?
You need to get ##\cos \theta ## as a function of ##x## or ##x## as a function of ##\theta## before you can integrate. Relate ##x, h, \theta##

erobz said:
You need to get ##\cos \theta ## as a function of ##x## or ##x## as a function of ##\theta## before you can integrate.
I am aware of that, but I wasn't able to find an efficient equation.

Heisenberg7 said:
I am aware of that, but I wasn't able to find an efficient equation.
Relate ##x, h## and ##\theta##( using trig). What do you get?

Heisenberg7
erobz said:
Relate ##x, h## and ##\theta##(trig). What do you get?
Well, ##\tan{\theta} = \frac{h}{x}##

Heisenberg7 said:
Well, ##\tan{\theta} = \frac{h}{x}##
So the ( a ) goal is to get ##\cos \theta## as a function of our current variable of integration ##x##. Decompose ##\tan \theta## into is constituent trig functions and just re-write that expression ( solve it for ##\cos \theta##)

Heisenberg7
erobz said:
So the ( a ) goal is to get ##cos \theta## as a function of our current variable of integration ##x##. Decompose ##\tan \theta## into is constituent trig functions and just re-write that expression ( solve it for ##cos \theta##)
$$\frac{\sin{\theta}}{\cos{\theta}} = \frac{h}{x} \implies \cos{\theta} = \frac{x\sin{\theta}}{h}$$

Heisenberg7 said:
$$\frac{\sin{\theta}}{\cos{\theta}} = \frac{h}{x} \implies \cos{\theta} = \frac{x\sin{\theta}}{h}$$
Now, can you see how to get rid of ##\sin \theta## in terms of the variable ##x##, and constant ##h## (think pythagoras) ?

Heisenberg7
erobz said:
Now, can you see how to get rid of ##\sin \theta## in terms of the variable ##x##, and constant ##h## (think pythagoras) ?
Oh, I see it now. So, $$\cos{\theta} = \frac{x\sin{\theta}}{h}\cos{\theta} = \frac{x \frac{h}{d}}{h} \implies \cos{\theta} = \frac{x}{d} \implies \cos{\theta} = \frac{x}{\sqrt{x^2 +h^2}}$$

erobz
Heisenberg7 said:
Oh, I see it now. So, $$\cos{\theta} = \frac{x\sin{\theta}}{h}\cos{\theta} = \frac{x \frac{h}{d}}{h} \implies \cos{\theta} = \frac{x}{d} \implies \cos{\theta} = \frac{x}{\sqrt{x^2 +h^2}}$$
Ok, sub it in to the integrand. Factor out the constants, and integrate.

Heisenberg7
erobz said:
Ok, sub it in to the integrand. Factor out the constants, and integrate.
And after integration, we get ##\sqrt{x_1^2 + h^2} - \sqrt{x_2^2 + h^2}##. And now I see that I did mess up my limits of integration. Anyway, tysm for the help!

Lnewqban
Heisenberg7 said:
And after integration, we get ##\sqrt{x_1^2 + h^2} - \sqrt{x_2^2 + h^2}##. And now I see that I did mess up my limits of integration. Anyway, tysm for the help!
You're welcome!

Heisenberg7

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