In procedure 2: suppose the inclined plane is tipped so it makes an angle = 11 d
ID: 1269113 • Letter: I
Question
In procedure 2: suppose the inclined plane is tipped so it makes an angle = 11 degrees with the horizontal. The 0 cm mark is at the bottom. The lower photogate is at the 35.1 cm mark, and the upper photogate is at the 70.8 cm mark along the track, Assume - The bottom of the track is at U = 0 J (potential energy is zero). - The total mass of the cart is 589 g. - The track is totally frictionless.
I calculated and got correct-->PE @ gate 1 is .39J and @ gate 2 it is .78J.
The question im stuck on-->
***Suppose the cart is released at rest at some point above the upper photogate. If the cart passes through the upper photogate at speed 0.574 m/s, at what speed will the cart pass through lower photogate?
Explanation / Answer
According to me, the simplest way to do this question is by the concept of Conservation of Energy.
That is, the potential energy lost from upper photogate to lower photogate = kinetic energy gained by the cart
So, Potential energy at upper photogate - P.E. at lower photogate = 1/2 m (v^2 - (0.574)^2)
0.78-0.39 = 0.589/2 (v^2 -0.33)
Solving , we get v^2 = 1.654
v = 1.286 m/s