Consider a roller coster car moving along a track from point A to point B to point C. a)if the roller coaster car starts from rest at point A (30m above the bottom of a dip (point b). What would be the speed of the car at the top of the next hill (point C), which is 20 m above the bottomm of the dip. Assume friction is negligible.
mgh=1/2mv^2
gh=1/2v^2
(9.8)(10) =1/2(v^2)
98=1/2v^2
v=14m/s
My issue is whether h is actually 20m (the distance b/w B and C)or is it 10m(the distance b/w A and C)
b) if instead the roller coaster car started with a speed of 20m/s at point A what is its speed at the top of the next hill (point C) Assume friction is negligible.
1/2mv_a^2=1/2mv_c^2
1/2v_a^2=1/2v_c^2
1/2(20)^2=1/2v_c^2
v_c=20 m/s
is this correct?
a) You got the right answer,but I would have worked it different, which answers your issue:
Total energy remains constant.
final energy= initial energy
PEfinal+KEfinal = PE initial + KE initial
mg20 + 1/2 mvf^2 = mg30 + 0
b) you did it wrong. Do it the way I outlined in a), you now have an initial KE.
For part a), the height (h) in the equation mgh = 1/2mv^2 represents the change in height between the initial position and the final position. In this case, the change in height would be the distance between point C and the bottom of the dip, which is 20m. So, you correctly used h = 20m in the calculations.
For part b), if the roller coaster car starts with a speed of 20m/s at point A, you can still use the equation 1/2mv_a^2 = 1/2mv_c^2, where v_a is the initial velocity and v_c is the velocity at point C.
Substituting the values, you get:
1/2(20)^2 = 1/2(v_c)^2
200 = 1/2(v_c)^2
(v_c)^2 = 400
v_c = √(400) = 20m/s
Therefore, your answer for part b) is correct as well.