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

Answer 2

The required value of b for the object to extend by 6.0cm is 63.93 Kg/s

The given parameters are:

Mass (m) = 2.5 kg

Stretch (y) = 6.0 cm

For the object to be at rest, then the vertical component of force acting on the object is 0.

This is represented as:

$\sum F_y = 0$

By response to damping, we have:

$ky -W = 0$

Where:

$W=mg$

So, we have:

$ky -mg = 0$

Substitute known values

$k*6.0cm -2.5kg * 9.81ms^{-2} = 0$

Express cm as m

$k*0.06m-2.5kg * 9.81ms^{-2} = 0$

$k*0.06m-24.525kgms^{-2} = 0$

Rewrite as:

$k*0.06m=24.525kgms^{-2}$

Divide through by 0.06m

$k=408.75Nm^{-1}$

Also, by critical damping, we have:

$b^2 =4km$

Make k the subject

$k = \frac{b^2}{4m}$

Substitute $k = \frac{b^2}{4m}$ in $k=408.75Nm^{-1}$

$\frac{b^2}{4m} = 408.75Nm^{-1}$

Substitute known value

$\frac{b^2}{4*2.5} = 408.75$

$\frac{b^2}{10} = 408.75$

Multiply both sides by 10

$b^2 = 4087.5$

Take the square root of both sides

$b = 63.93$

Hence, the required value of b for the object to extend by 6.0cm is 63.93 Kg/s

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