A ski starts from rest and slides down a 25° incline 100 m long. Use energy methods.
(a) If the coefficient of friction is 0.09, what is the ski s speed at the base of the incline?
(b) If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level?
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Answers & Comments
The height of the incline is 100 sin 25 = 42.26 m
PE at the top is mgh = m9.8• 42.26 = m414.2 J
Force of friction = µmg cos 25 = m0.7994 N
Energy used by friction = Fd = m79.94 J
KE at bottom = m414.2 – m79.94 = m334.3 J
KE = ½mV² = m334.3
V² = 2•334.3
V = 25.0 m/s
Force of friction is 0.09mg
a = f/m = 0.09mg/m = 0.09g = 0.882 m/s²
v = v₀ + √(2ad)
2ad = V²
d = V²/2a = 2•334.3/2• 0.882 = 379 m