Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 29 ∘ angle. The block's initial speed is 9.0 m/s . The coefficient of kinetic friction of wood on wood is μk=0.200.

A)What vertical height does the block reach above its starting point?

B)What speed does it have when it slides back down to its starting point?

Relevance

A) initial KE becomes friction work and PE:

½mv² = µmgdcosΘ + mgdsinΘ

mass m cancels

½v² = d*g*(µcosΘ + sinΘ)

½(9.0m/s)² = d * 9.8m/s² * (0.200*cos29º + sin29º)

solves to

d = 6.26 m

and so the height is

h = 6.26m*sin29º = 3.0 m

B) The work done by friction for the back-and-forth route is

W = 2 * 2.0kg * 9.8m/s² * 0.200 * 6.26m * cos29º = 42.9 J

initial KE = ½ * 2.0kg * (9.0m/s)² = 81 J

final KE = initial KE - Work = 38.1 J = ½ * 2.0kg * V²

so the final speed is

V = 6.2 m/s

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• A)What vertical height does the block reach above its starting point?

The initial kinetic energy will convert to potential energy and and friction work

½mv² = mgh + μNd = mgh + μmgcosθ(h/sinθ)

½v² = gh(1 + μcotθ)

h = v²/2g(1 + μcotθ)

h = 9.0² / 2(9.8)(1 + 0.200cot29) = 3.0369... ≈ 3.0 m

B)What speed does it have when it slides back down to its starting point?

The initial potential energy will convert to kinetic energy and friction work

mgh = ½mv² + μmgcosθ(h/sinθ)

gh -  μgcosθ(h/sinθ) = ½v²

v² = 2gh(1 - μcotθ) = 2(9.8)(3.0369)(1 - 0.200cot29) = 38.04678...

v = 6.1682...≈ 6.2 m/s

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