# In the given figure initially spring is in its natural length and system is released from rest. Finally block again comes to rest at mean position due to air resistance. Find the total work done by?

## 1) gravitational force 2) spring force 3) force due to air resistance

Aug 2, 2018

This is what I get

#### Explanation:

When the system is released from rest, the mass $M$ is pulled down by gravity with a force

${F}_{G} = M g$
where $g$ is acceleration due to gravity.

This force acts on the spring (assumed mass-less) which initially is in natural unstreched position. Let the elongation of the spring be ${y}_{0}$ where gravitational force is balanced by the restoring force of the spring. This is system's new equilibrium position. Here we have the expression

$M g = k {y}_{0}$
$\implies {y}_{0} = \frac{M g}{k}$ .....(1)

At the mean position of the system, taken as origin of coordinates, let mass have velocity $v$. It shoots past this position in the downwards direction. Thereafter, the restoring force of the spring takes over and pulls back the mass in the upwards direction. The system keeps on oscillating and finally comes to rest at the new equilibrium position due to air resistance losses.

1. Initial total energy equation of the system ${E}_{i} = P {E}_{\text{spring"+GPE_"Mass"+KE_"Mass}}$
$\implies {E}_{i} = 0 + M g {y}_{0} + 0$
$\implies {E}_{i} = M g {y}_{0}$
2. Final total energy equation of the system
${E}_{f} = P {E}_{\text{spring"+GPE_"Mass"+KE_"Mass"+E_"air friction}}$
$\implies {E}_{f} = \frac{1}{2} k {y}_{0}^{2} + 0 + 0 + {E}_{\text{air friction}}$
$\implies {E}_{f} = \frac{1}{2} k {y}_{0}^{2} + {E}_{\text{air friction}}$
Using Law of conservation of energy we get
$M g {y}_{0} = \frac{1}{2} k {y}_{0}^{2} + {E}_{\text{air friction}}$
$\implies {E}_{\text{air friction}} = M g {y}_{0} - \frac{1}{2} k {y}_{0}^{2}$

Work done by

1. gravitational force $= M g {y}_{0}$
2. spring force $= \frac{1}{2} k {y}_{0}^{2}$
3. force due to air resistance $= M g {y}_{0} - \frac{1}{2} k {y}_{0}^{2}$
where ${y}_{0}$ is given by (1).
Aug 3, 2018

Definition of work:

$\delta W = \boldsymbol{F} \cdot \delta \boldsymbol{r}$

Assuming this question is a reference to an (under) damped oscillation, it is helpful for a moment to ignore the damping (air resistance, friction, whatever dissipative force is at play).

• Over an undamped oscillation, gravity will do zero new work on the suspended mass. Gravity is ${\boldsymbol{F}}_{g} = m g \setminus \boldsymbol{\hat{y}}$, it always acts in the downward positive y-direction, so it variously encourages and opposes motion ie does positive or negative work.

• Over half an undamped oscillation, the spring will do zero new work on the suspended mass. The spring is restorative, always acting toward the centre of the oscillation: ${\boldsymbol{F}}_{s} = - k y \setminus \boldsymbol{\hat{y}}$. Again the spring variously encourages and opposes motion ie does positive or negative work on the suspended mass.

Adding the damping will not change this. It is the damping that extracts the energy from the mass-spring system. The physics is simple. As the mass collides with the air molecules, its energy is dissipated. The mass does work on its surroundings, or if you like, the air resistance does negative work on the mass-spring.

With a judiciously chosen damping term, and positive y downwards:

$F \left(y , \dot{y}\right) = - k y + m g - {\underbrace{m \nu \dot{y}}}_{\text{damping " propto " velocity}}$

• $\implies m \ddot{y} + k y + m \nu \dot{y} - m g = 0 q \quad \triangle$

The total energy $E$ in the system is:

$E = {\underbrace{\frac{1}{2} k {y}^{2}}}_{\text{spring") + underbrace(1/2 m dot y^2)\_("kinetic") - underbrace( m gy )\_("gravitational}}$

$\frac{\mathrm{dE}}{\mathrm{dt}} = k y \dot{y} + m \dot{y} \ddot{y} - m g \dot{y}$

$= \left(k y + m \ddot{y} - m g\right) \dot{y}$

Which from $\triangle$:

$= - \left(m \nu \dot{y}\right) \dot{y} = - m \nu {\dot{y}}^{2}$

Solving $\triangle$ for the underdamped oscillator and working out the impact of non-dissipative forces over the oscillations is fiddly, but very do-able, and will confirm the big-picture approach.