In procedure 2: suppose an inclined plane is 120 cm long, and is tipped at angle

Question

In procedure 2: suppose an inclined plane is 120 cm long, and is tipped at angle ? = 10.5o. The 0 cm mark is at the bottom. (NOTE: In the lab, the 0 cm mark will be at the top.) Photogate 1 is at the 35.6 cm mark, and photogate 2 is at the 64.8 cm mark along the track, Assume - The bottom of the track is at 0 J potential energy. - The total mass of the cart is 131 g. - The track is totally frictionless a) Find the potential energy of the cart . - At photogate 1: J - At photogate 2: J b) Suppose the cart is released at rest at photogate 2 (the higher point on the track). Find the speed of the cart as it passes photogate 1. m/s

Explanation / Answer

PE (35.6 mark)= 0.131*9.8*.356*sin 10.5 = 0.0833 J

PE (64.8 mark) = 0.131*9.8*.648*sin 10.5 = 0.1516 J

0.5*0.131*v^2 = .1516-.0833

v = 1.02 m/s

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