Answer:
Change in potential energy of the block-spring-Earth
system between Figure 1 and Figure 2 = 1 Nm.
Explanation:
Here, spring constant, k = 50 N/m.
given block comes down eventually 0.2 m below.
here, g = 10 m/s.
let block be at a height h above the ground in figure 1.
⇒In figure 2,
potential energy of the block-spring-Earth
system = m×g×(h - 0.2) + 1/2× k × x². where, x = change in spring length.
⇒ Change in potential energy of the block-spring-Earth
system between Figure 1 and Figure 2 = (m×g×(h - 0.2)) - (1/2× k × x²)
= (1×10×0.2) - (1/2×50×0.2×0.2) = 1 Nm.
1 kg = 1000 gr
6.4 kg = 6400 gr
ill try to get back to you, im doing the test now
Answer:
W = 1562.5 J
Explanation:
Path 1:
W₁ = F₁*d₁ = 385 N * 2.5 m = 962.5 J
Path 2:
W₂ = F₂*d₂ = 130 N * 10 m = 1300 J
Path 3:
W₃ = F₃*d₃ = (-350 N) * 2 m = - 700 J (opposite to the motion)
We get
W = W₁ + W₂ + W₃ = 962.5 J + 1300 J + (- 700 J) = 1562.5 J
Answer:
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