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A 750 kg roller coaster car drops from rest at a height of 90 m along a frictionless track. What is velocity of the roller coaster at the top of a second hill that is 60 m high? What is the average force required to make it stop along a 120 m stretch of flat track at ground level.

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MathPhys

Answer:

24.2 m/s

5510 N

Explanation:

Energy is conserved. Initial potential energy = final potential energy + final kinetic energy.

PE₀ = PE + KE

mgh₀ = mgh + ½ mv²

gh₀ = gh + ½ v²

v = √(2g (h₀ − h))

v = √(2 × 9.8 m/s² (90 m − 60 m))

v = 24.2 m/s

Work = change in energy

W = ΔE

Fd = mgΔh

F (120 m) = (750 kg) (9.8 m/s²) (90 m)

F = 5510 N

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