The strain energy stored in a linear spring is
SE = (1/2)*k*x²
where
k = the spring constant
x = the extension (or compression) of the spring
Given:
k = 470 N/m
x = 17.0 cm = 0.17 m
Therefore
SE = 0.5*(470 N/m)*(0.17 m)² = 6.7915 J
Answer: 6.8 J (nearest tenth)
Answer:
60 000 N
Explanation:
1 pa = 1 N/m^2
you have 300 000 of these = 300 000 N /m^2
but only an area of .2 m^2
300 000 N / m^2 * .2 m^2 = 60 000 N
Answer:
b ≈ 64 Kg/s
Explanation:
Given
Fd = −bv
m = 2.5 kg
y = 6.0 cm = 0.06 m
g = 9.81 m/s²
The object in the pan comes to rest in the minimum time without overshoot. this means that damping is critical (b² = 4*k*m).
m is given and we find k from the equilibrium extension of 6.0 cm (0.06 m):
∑Fy = 0 (↑)
k*y - W = 0 ⇒ k*y - m*g = 0 ⇒ k = m*g / y
⇒ k = (2.5 kg)*(9.81 m/s²) / (0.06 m)
⇒ k = 408.75 N/m
Hence, if
b² = 4*k*m ⇒ b = √(4*k*m) = 2*√(k*m)
⇒ b = 2*√(k*m) = 2*√(408.75 N/m*2.5 kg)
⇒ b = 63.9335 Kg/s ≈ 64 Kg/s
