Ask Question
17 January, 13:43

A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.

+2
Answers (1)
  1. 17 January, 17:27
    0
    A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.

    how much work is done on the monitor by (a) friction, (b) gravity

    work (friction) = 453.5J

    work (gravity) = - 453.5J

    Explanation:

    Given that,

    mass = 14kg

    displacement length = 5.50m

    displacement angle = 36.9°

    velocity = 2.30cm/s

    F = ma

    work (friction) = mgsinθ. displacement

    = (14) (9.81) (5.5sin36.9°)

    = 453.5J

    work (gravity)

    = the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)

    = 126.9°

    work (gravity) = (14) (9.81) (5.5cos126.9°)

    = - 453.5J
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers