2.A roller coaster of mass 500kg approaches the bottom of a loop at a speed 25ms-1. the loop can be approximated as a circle with radius 5.0m
(a) Calculate the kinetic energy of the roller coaster at the bottom of the loop
(b)Calculate the speed of the roller coaster when it reaches the top of the loop
3. A frictionless hydraulic press is used to lift a car. The area of piston X is 0.80m2 while the area of piston Y is 10m2. The total mass of the call in the piston Y is 1200 KG. Take g=10m s-2
(a) calculate the minimum force exerted by piston X to lift the car
(b) calculate the distace moved by piston X if the car is lifted by 0.40 m
•Lastly i cant understand these formulas in ur blog, could you pls abbrivate the 2,3 and 4th formula. It will be helpful to me....
Pressure = Force/Area
Pressure = Force/Area
•=Weight/Area
•=mg/A
•=pVg/A
•= pAhg/A
•= hpg
Answers
ReplyDelete(1) K.E gain = Gravitational P.E. loss
0.5 m v^2 = mgh
v = square root of (2gh)
v = square root of (2x10x5.0)= 10.0 m/s
(2) K.E. at the bottom of the loop
= 0.5 x m x v^2
= 0.5 x 500 x 25 x 25 = 156250 J
= 156 000 J to 3 s.f.
(3) radius of circular loop is 5.0 m, so diameter = 10.0 m.
Gravitational P.E.gained = mgh = 500 x 10 x 10.0
= 50 000 J
K.E. left = 156 000 - 50 000 J = 106, 000 J
0.5 x 500 x v^2 = 106, 000
v^2 = (2x106,000)/500
v = square root of (2x106,000/500)
= 20.59 m/s = 21 m/s (to 2 s.f.)
Pressure at X = Pressure at Y
F/0.80 = 1200 x 10 / 10
F = 960 N
Minimum force exerted by the piston at X to raise the car = 960 N
Work done by the piston = Work done on the car
960 x distance moved = 1200 x 10 x 0.40
distance moved by piston = 1200 x 4/960 = 5.0 m