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I had a lab that tested the dependence of gravitational potential energy on its position and the goal out of each exercise was to see if kinetic energy equaled potential energy. A cart was on a flat frictionless track, with a string attached to it and a hanging 20-g mass on the opposite end of the string hanging down by a pulley. We saw that the kinetic energy was inversely proportional to the PE in this exercise and that in this case there were multiple kinetic energy forms and a single potential energy term.

My confusion lied in the exercise that followed, where the track was now raised at an angle, the front of the cart pointing down the ramp. The difference aside from the elevated track was now we had multiple potential and kinetic energy forms. Our lab instructor mentioned that the carts height will change as it moves along the ramp but NOT by the same amount as the hanging mass. I don't understand that statement because when I plotted the potential and kinetic energy vs time graphs for both exercises they were basically identical, showing that both KE and PE were inversely proportional. Am I not understanding the question correctly when I relate it to the the kinetic energy and potential energy graphs?

Disclaimer: I know the colors suck and are distracting but the software I use in class to collect data doesn't have an easy way of making the graphs appealing to the eye.

For both exercises I used the equation to plot the graphs where kinetic energy was:

$KE=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$

and potential energy was:

$PE = mgh_i+mgh_f$

Flat Surface Exercise

Flat Surface Exercise

Surface Raised At Angle Exercise

Elevated Surface Exercise

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  • $\begingroup$ Are you sure they said direction and not say distance or height? This is straightforward to prove that the inclined carts change in height is not the same as the falling mass just using trigonometry. $\endgroup$
    – Triatticus
    Commented Oct 11, 2021 at 19:58
  • $\begingroup$ Oof I'm dumb I'll edit that, I just went back and it did say height not direction. That's the part I'm stuck on is how I'd use the trig components to solve this. $\endgroup$
    – Iseez Ice
    Commented Oct 11, 2021 at 20:05
  • $\begingroup$ Your best bet is to start with a drawing of the set-up $\endgroup$
    – Triatticus
    Commented Oct 11, 2021 at 20:08

1 Answer 1

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Where the cart is on a flat surface, when you release the weight it drops and both it and the cart gain speed, which means that they gain kinetic energy. Since they are connected by a string, they both move at the same speed. Assuming they are stationary at the start of the experiment, the kinetic energy they gain is half of their combined mass times the square of their velocity at the point you measure it. You can equate that to mgh, where m is the mass of the weight and h is the vertical distance through which it has dropped. The cart is moving horizontally, so its potential energy does not change.

Then the cart is on a slope, it is no longer moving horizontally, but gradually moving downwards. It does not move downwards as quickly as the weight, since the weight is falling vertically. The gain in kinetic energy can be calculated in the same way as before, namely as half the combined mass of the cart plus the weight times the square of their common speed. The PE lost by the weight is again mgh, where m is the mass of the weight and h is the height through which it has fallen. What you now have to calculate is the loss of PE of the cart. You know that the distance it has travelled along the slope is equal to h, since the cart and the weight are connected by a string. So now you need only resolve that into two components, namely the horizontal component and the vertical. You should find that the vertical component is h times the sine of the angle the slope makes with the horizontal.

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  • $\begingroup$ So what I ended up getting was GPE = mg(0.93-x)sin(2). Where 0.93 was the distance the cart travelled from it's initial point to it's final point. $\endgroup$
    – Iseez Ice
    Commented Oct 11, 2021 at 23:04
  • $\begingroup$ Also the 2 is the degree the ramp was raised $\endgroup$
    – Iseez Ice
    Commented Oct 11, 2021 at 23:59

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