Answer:
a) Δy = 0.144 m
b) W = 0.145 J
c) Us = 0.32 J
d) ymax = 0.144 m
Explanation:
a) First let's find the spring constant using Hooke's Law
F = k*Δy ⇒ k = F/Δy
where
F = m*g = 0.1 kg*9.81 m/s² = 0.981 N
and Δy = 0.07 m. Hence
k = 0.981 N/0.07 m = 14.014 N/m ≈ 14 N/m
In order to find the position of the block when we let it go, we need to find the force that caused this expansion in the spring, we know that the reading of the scale was 3 N and this reading includes the force we want to find and the weight of the block, therefore:
f = 3 N - F = 3 N - 0.981 N = 2.019 N
Now that we have found the force we can use Hooke's Law in order to find the position of the block
f = k*Δy ⇒ Δy = f/k
⇒ Δy = 2.019 N/14 N/m
⇒ Δy = 0.144 m
b) First, notice that there are two kind of potential energy: the potential energy in the spring and the potential energy due to the gravitational field:
W = ΔU
W = ΔUs + ΔUg
W = (Usf - Usi) + (Ugf - Ugi)
Notice that
Us = 0.5*k*y²
where
yf = 0.07 m + 0.144 m = 0.214 m and
yi = 0.07 m
and we will take the zero level to be the equilibrium position where the block was hanging at rest. Hence
W = 0.5*k*(yf² - yi²) + m*g*(0 - Δy)
⇒ W = 0.5*14 N/m*((0.214 m)² - (0.07 m)²) + (0.1 kg)*(9.81 m/s²)*(0 - 0.144 m)
⇒ W = 0.145 J
c) When we let the block go the spring was stretched by
y = 0.07 m + 0.144 m = 0.214 m
Therefore:
Us = 0.5*k*y²
⇒ Us = 0.5*14 N/m*(0.214 m)²
⇒ Us = 0.32 J
d) Because the position that we pulled the block to it is considered as the amplitude for the vibrational motion that will happen after we release the block, then the maximum height the particle will reach above the equilibrium position is
ymax = Δy = 0.144 m
