A 60.0 kg skier starts from rest at the top of a ski slope 65.0 m high. (a) If friction forces do -10.5 kJ of work on her as she descends, how fast is she going at the bottom of the slope

-2.69m/s²

Explanation:

Workdone = Force × distance

Workdone = mass × acceleration × distance

Given

Mass = 60.0kg

Workdone = -10.5kJ = -10500J

Distance = 65.0m

Acceleration = ?

Substitute the given parameters into the formula and get acceleration

-10500 = 60(a)(65)

-10500 = 3900a

a = -10500/3900

a = -105/39

a = -2.69m/s²

Hence she is going at -2.69m/s² at the bottom of the slope.

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explanation: