The law applied here is Hooke's Law which describes the force exerted by the spring with a given distance. The equation for this is F = kΔx, where F is the force in Newtons, k is the spring constant in N/m while Δx is the displacement in meters.
If you want to find work done by a spring, this can be solved by using differential equations. However, derived equations are already ready for use. The equation is
W = k[{x₂-x₁)² - (x₁-xn)²],
where
xn is the natural length
x₁ is the stretched length
x₂ is also the stretched length when stretched even further than x₁
In this case xn =x₁. So, that means that (x₁-xn) = 0 and (x₂-x₁) = 11 cm or 0.11 m.
Then, substituting the values,
2 J = k (0.11² -0²)
k = 165.29 N/m
Finally, we use the value of k to the Hooke's Law to determine the Force.
F = kΔx = (165.29 N/m)(0.11 m)
F = 18.18 Newtons
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
D. because they aren't mixed up evenly
Answer:
Inside a femur bone found in the woods,
saliva of gum
Answer:
Electric field intensity is the force experienced by a test charge q in a electric field E.