Question

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

Answer 1

Answer:

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