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# An extreme skier, starting from rest, coasts down a mountain that makes an angle of 30.9 ° with the horizontal. The coefficient of kinetic f

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An extreme skier, starting from rest, coasts down a mountain that makes an angle of 30.9 ° with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.241. She coasts for a distance of 19.0 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.76 m below the edge. How fast is she going just before she lands?

physics
Oct 20, 2014

#1
+5

1.  On the cliff.

Resolving forces parallel to the surface we have m.g.sinθ - μ.m.g.cosθ down the slope where θ is 30.9° and μ is 0.241;  m is mass and g ia gravitational acceleration.

From Newton's 2nd law of motion this must equal m.a where a is (constant) acceleration.

This means a = g.(sinθ - μ.cosθ)

Hence velocity, v, at the edge of the cliff is v = √(2.a.s)  where s is 19 metres.

2.  Off the cliff.

The horizontal velocity, vh, is constant (ignoring air resistance etc.) at  vh = v.cosθ.

The initial vertical velocity, vv, is vv = v.sinθ.

For the vertical motion we have vvf2= vv2 + 2.g.sv  where vvf is final vertical velocity and sv is 3.76m.

3. Landing

The magnitude of the resultant velocity, vr, when she lands is given by:  vr = √(vvf2 + vh2

You can insert the numbers.

.

Oct 21, 2014

#1
+5

1.  On the cliff.

Resolving forces parallel to the surface we have m.g.sinθ - μ.m.g.cosθ down the slope where θ is 30.9° and μ is 0.241;  m is mass and g ia gravitational acceleration.

From Newton's 2nd law of motion this must equal m.a where a is (constant) acceleration.

This means a = g.(sinθ - μ.cosθ)

Hence velocity, v, at the edge of the cliff is v = √(2.a.s)  where s is 19 metres.

2.  Off the cliff.

The horizontal velocity, vh, is constant (ignoring air resistance etc.) at  vh = v.cosθ.

The initial vertical velocity, vv, is vv = v.sinθ.

For the vertical motion we have vvf2= vv2 + 2.g.sv  where vvf is final vertical velocity and sv is 3.76m.

3. Landing

The magnitude of the resultant velocity, vr, when she lands is given by:  vr = √(vvf2 + vh2

You can insert the numbers.

.

Alan Oct 21, 2014