a roller coaster goes over a peak in the track which is 60.0 m high, with a speed of 14.0 m/s. what is its speed at the next trough which is 20.0 m high?



Answer :

Answer:31.6m/s

Explanation:

Conservation of Energy

mgh+1/2mv^2=mgh+1/2mv^2

The masses cancel out

gh+1/2v^2=gh+1/2v^2

g=10m/s^2

(10m/s^2)(60m)+(1/2)(14m/s)^2=(10m/s^2)(20m)+1/2v^2

600+98=200+.5v^2

698=200+.5v^2

498=.5v^2

v^2=996

v=31.6m/s

Answer:

31.3 m/s

Explanation:

Energy is conserved. The roller coaster starts at the peak with both gravitational potential energy (PE) and kinetic energy (KE). When it reaches the next trough, some of the initial potential energy is converted to kinetic energy.

PE₀ + KE₀ = PE + KE

mgh₀ + ½ mv₀² = mgh + ½ mv²

gh₀ + ½ v₀² = gh + ½ v²

v² = 2g (h₀ − h) + v₀²

v² = 2 (9.8 m/s²) (60.0 m − 20.0 m) + (14.0 m/s)²

v = 31.3 m/s

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