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

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

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,

Therefore, the additional work needed is

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This gots to be the answer, average, seawater in the world's oceans has a salinity of approximately 3.5%, or 35 parts per thousand.
Answer:
Explanation:
Given that,
Mass of sledge hammer;
Mh =2.26 kg
Hammer speed;
Vh = 64.4 m/s
The expression fot the kinetic energy of the hammer is,
K.E(hammer) = ½Mh•Vh²
K.E(hammer) = ½ × 2.26 × 64.4²
K.E ( hammer) = 4686.52 J
If one forth of the kinetic energy is converted into internal energy, then
ΔU = ¼ × K.E(hammer)
∆U = ¼ × 4686.52
∆U = 1171.63 J
Thus, the increase in total internal energy will be 1171.63 J.