Work needed: 720 J
Explanation:
The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by
![E=\frac{1}{2}kx^2](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
where
k is the spring constant
x is the stretching of the spring from the equilibrium position
In this problem, we have
E = 90 J (work done to stretch the spring)
x = 0.2 m (stretching)
Therefore, the spring constant is
![k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2E%7D%7Bx%5E2%7D%3D%5Cfrac%7B2%2890%29%7D%7B%280.2%29%5E2%7D%3D4500%20N%2Fm)
Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of
x = 0.2 + 0.4 = 0.6 m
Substituting,
![E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J](https://tex.z-dn.net/?f=E%27%3D%5Cfrac%7B1%7D%7B2%7Dkx%5E2%3D%5Cfrac%7B1%7D%7B2%7D%284500%29%280.6%29%5E2%3D810%20J)
Therefore, the additional work needed is
![\Delta E=E'-E=810-90=720 J](https://tex.z-dn.net/?f=%5CDelta%20E%3DE%27-E%3D810-90%3D720%20J)
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Answer:
x₂ = 16 g m₂ / k
Explanation:
The spring in all cases must comply with Hooke's law
F = k x
Newton's diagram for equilibrium has
F - W = 0
k x₁ = m₁ g
k = g m₁ / x₁
When the elevator moves you have a clarification
F-W = m a
F = m (g + a)
Compression is
K x₂ =4 m (g + 3g)
x₂ = 4m / k (4g)
x₂ = 16m2 / k g
The high temperature of the heated air inside the balloon causes the air particles to spread out. This leads to a lower density compared to colder air surrounding the balloon.
Therefore the low density causes the balloon to rise up.
Yes you are :)
hope i helped