# Help please?

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?

### 2 Answers

- NCSLv 71 month agoFavorite Answer
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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- WhomeLv 71 month ago
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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