Answer:
d. The scale's resolution is too low to read the change in mass
Explanation:
If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:
F = kx
since,
w = Fd
dw = Fdx
integrating and using value of F, we get:
ΔE = (0.5)kx²
where,
ΔE = Energy added to spring
k = spring constant
x = displacement
The spring constant is typically in range of 4900 to 29400 N/m.
So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:
ΔE = (0.5)(29400 N/m)(1 m)²
ΔE = 1.47 x 10⁴ J
Now, if we convert this energy to mass from Einstein's equation, we get:
ΔE = Δmc²
Δm = ΔE/c²
Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²
<u>Δm = 4.9 x 10⁻¹³ kg</u>
As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.
Since, the scale only gives the mass value upto 1 decimal place.
Thus, it can not determine such a small change. So, the correct option is:
<u>d. The scale's resolution is too low to read the change in mass</u>
