promotion image of download ymail app
Promoted
Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

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

Relevance
  • NCS
    Lv 7
    1 month ago
    Favorite 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

    If you find this helpful, please select Favorite Answer!

    • Commenter avatarLogin to reply the answers
  • Whome
    Lv 7
    1 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

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.