Roller Coaster Physics Problem?
Suppose a roller coaster (h at point1 = 35 m, h at point 2 = 14 m, h at point 3 = 25) passes point 1 with a speed of 1.10 m/s.
If the average force of friction is equal to one third of its weight, with what speed will it reach point 2?
The distance traveled is 45.0 m.
Comments
OK, in the S = 45 m, the RC loses QE = FS = 1/3 mgS of its energy between 1 and 2.. mg = W is the weight of the RC.
At 1, TE = ke + pe = 1/2 mv^2 + mgH; where H = 35 m and v = 1.1 mps. At 2, TE = 1/2 mV^2 + mgh - QE; where h = 14 m.
From the conservation of energy, we have 1/2 mv^2 + mgH = 1/2 mV^2 + mgh - 1/3 mgS
The m's cancel out; so we have
v^2 + 2gH = V^2 + 2gh - 2/3 gS and solve for V = ? mps, the speed at point 2.
V = sqrt(v^2 + 2gH - 2gh + 2/3 gS) = sqrt(1.1^2 + 2*9.8*(35 - 14) + (2/3)*9.8*45) = 26.58589852 = 26.6 mps. ANS.
oh i'm getting it, the time it would want to take for it to bypass up the (s) first hill is nice and speed, it would want to take 16 seconds to bypass down the second one hill and advance speed and time till it is going up yet another hill is expected good will take a lot less or extra seconds to bypass up/down the second one hill and others and advance/shrink going up/down the hills utilising newtons guidelines of one million, 2, and three a million is at the same time as an merchandise or shifting merchandise stops or strikes and volume 2 is the route in at the same time as the article strikes and volume 3 is utilising action/reaction situation like a vehicle hits yet another the first vehicle exerts skill on the second one and the second one exerts it decrease back
google it :-)