Question

A spring balance consists of a pan that hangs from a spring. A damping force Fd = −bv is applied to the balance so that when an object is placed in the pan it comes to rest in the minimum time without overshoot. Determine the required value of b for an object of mass 2.5 kg that extends the spring by 6.0 cm. (Assume g = 9.81 m s−2.) 2.

Answer 1

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