Starting from rest, a 30.8 kg child rides a 9.75 kg sled down a frictionless ski slope. At the bottom of the hill, her speed is 6.0 m/s . If the slope makes an angle of 15.1 ° with respect to the horizontal, how far along the hill did she slide on her sled?

- We will apply the conservation of energy principle on the system consisting of child and sledge. The change in potential energy ΔP.E of the system from top to bottom gives rise to change in kinetic energy ΔK.E.

ΔP.E = ΔK.E

(Mc + Ms)*g*h = 0.5*(Mc + Ms)*(vf^2 - vi^2)

Where, h: the change in vertical height

h = 0.5*(vf^2 - vi^2) / g

h = 0.5*vf^2 / g

h = 0.5*6.0^2 / 9.81

h = 1.835 m

- The distance traveled along the slope (s) can be calculated by trigonometric ratio:

sin ( θ ) = h / s

s = h / sin ( θ )

s = 1.835 / sin ( 15.1 )

s = 7.044 m

andromache andromache · Answer 2 · 2022-03-01T16:13:04+01:00

The distance that along the hill did she slide on her sled is s = 7.044 m

Conversation of energy:

Since The mass of child Mc = 30.8 kg, The mass of sledge Ms = 9.75 kg, The slope θ = 15.1°, The speed of of sledge at bottom vf = 6.0 m/s, and The initial velocity at top vi = 0

Now here we apply the convesation of energy

ΔP.E = ΔK.E

(Mc + Ms)*g*h = 0.5*(Mc + Ms)*(vf^2 - vi^2)

Here h be the change in vertical height

So,

h = 0.5*(vf^2 - vi^2) / g

h = 0.5*vf^2 / g

h = 0.5*6.0^2 / 9.81

h = 1.835 m

Now the distance traveled is

sin ( θ ) = h / s

s = h / sin ( θ )

s = 1.835 / sin ( 15.1 )

s = 7.044 m

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