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22 April, 22:38

Dźwig podniósł kontener o masie m = 80 kg na wysokość h = 10 m. Pierwsze 5 m kontener przebył z przy-

spieszeniem o wartości a = 1 m/s2, a następne 5 m - ruchem jednostajnym. Przyjmij, że g = 10 m/s2. Oblicz:

a) całkowitą pracę wykonaną przez dźwig,

b) ile razy praca wykonana podczas przyspieszania kontenera jest większa od pozostałej pracy.

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  1. 23 April, 01:20
    0
    a) W_total = 8240 J, b) W₁ / W₂ = 1.1

    Explanation:

    In this exercise you are asked to calculate the work that is defined by

    W = F. dy

    As the container is rising and the force is vertical the scalar product is reduced to the algebraic product.

    W = F dy = F Δy

    let's apply this formula to our case

    a) Let's use Newton's second law to calculate the force in the first y = 5 m

    F - W = m a

    W = mg

    F = m (a + g)

    F = 80 (1 + 9.8)

    F = 864 N

    The work of this force we will call it W1

    We look for the force for the final 5 m, since the speed is constant the force must be equal to the weight (a = 0)

    F₂ - W = 0

    F₂ = W

    F₂ = 80 9.8

    F₂ = 784 N

    The work of this fura we will call them W2

    The total work is

    W_total = W₁ + W₂

    W_total = (F + F₂) y

    W_total = (864 + 784) 5

    W_total = 8240 J

    b) To find the relationship between work with relate (W1) and work with constant speed (W2), let's use

    W₁ / W₂ = F y / F₂ y

    W₁ / W₂ = 864/784

    W₁ / W₂ = 1.1
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