suppose an inclined plane is 120 cm long, and is tipped at angle ? = 10.5 o . Th

Question

suppose an inclined plane is 120 cm long, and is tipped at angle ? = 10.5o. The 0 cm mark is at the bottom.

Photogate 1 is at the 22.8 cm mark, and photogate 2 is at the 58.1 cm mark along the track, Assume
- The bottom of the track is at 0 J potential energy.
- The total mass of the cart is 789 g.
- The track is totally frictionless

The potential energy of

- At photogate 1: 1 J
- At photogate 2: 2 J

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?

please explain step by step

Explanation / Answer

PE = mgh = mgl*cosx PE1 at gate 1 = 0.789*9.8*(0.228*sin10.5) = 0.322 J and PE2 at gate 2 = 0.789*9.8*(0.581*sin10.5) = 0.82 J b) PE turns to KE mgh = 1/2mv^2 v = sqrt(2g(h2 - h1)) = sqrt(2g(l2-l1)cosx) v = sq rt[2*9.8*(0.581-0.228)*sin 10.5] = 1.124 m/s

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.