$K=\frac{1}{2} m v^2$

If a particle moves from position 1 to position 2, such that at position 1 its speed is $v_i$, and at position 2 the speed is $v_f$.

Kinetic energy at 1 = $\frac{1}{2} m v_i^2$

Kinetic energy at position 2 $=\frac{1}{2} m v_f^2$

$\Delta$K $=\frac{1}{2} m v_f^2-\frac{1}{2} m v_i^2$

$\Delta$ $\rightarrow$ change in final state - initial state.

$W$ = $K_f$ – $K_i$ = $\Delta K$

W is work done by external forces acting on the particle as it moves from state 1 to state 2.

Work done by external forces will be the work done by net external forces.

$\sum \vec{F}_{i \text { ext. }}$.

$K=\frac{1}{2} m v^2$

Differentiate with respect to time,

$\frac{d K}{d t}=\frac{1}{2} m \frac{d}{d t}\left(v^2\right)=\frac{1}{2} m 2v \frac{d v}{d t} = m\frac{d v}{d t} v$

$\frac{d K}{d t} = F v$

$v \rightarrow \frac{d x}{d t}$

$\frac{d K}{d t} = F\frac{d x}{d t}$

$\int_i^f d k=\int_{x_i}^{x_f} F d x$

$\Delta K=W$

Force is along one direction F $\rightarrow x$.

The work energy theorem is valid for 2D or 3D motion.

$K=\frac{1}{2} m(\vec{v} \cdot \vec{v})$

$\frac{d K}{d t}=\frac{1}{2} m 2\left(\vec{v} \cdot \frac{d \vec{v}}{d t}\right)=\vec{v} \cdot m \frac{d \vec{v}}{d t}$

$\frac{d K}{d t}=\vec{F} \cdot \vec{v}$

$\text { for a particle } \vec{v}=\frac{d \vec{r}}{d t}$

$\frac{d K}{d t}=\vec{F} \cdot \frac{d \vec{r}}{d t}$

$\int d K=\int \vec{F} \cdot d \vec{r}$

Work kinetic energy theorem is an integrated form of Newton's second law.

$F=m \frac{d v}{d t}$

$F=m\frac{ d v}{d x} \frac{d x}{d t}=m \frac{d v}{d x} v$

$\int F d x=\int m v{ d v} = m\left[\frac{v^2}{2}\right]_i^f$

Valid only if $\vec{a}, v, \vec{r}$ are being measured w.r.t. an inertial frame of reference.

The normal reaction acts perpendicular to the surface.

Will not do any work.

A particle moves on a circular path.

The particle is moving in a direction which is perpendicular to the string force.

Throw the ball up in air from ground with speed v.

As the ball moves up gravity starts acting down, a point comes where v = 0.

At top position (highest)

v = 0, K = 0

Now gravity increases the speed as the ball comes down,so kinetic energy increases.

Gravity is doing work on the ball throughout.

In the upward motion

The work done by gravity < 0

$\Delta$K = W

At the top position kinetic energy becomes 0.

Downward motion

Gravity is acting down and the displacement vector $\vec{r}$ is also down. So the work done is positive. w > 0

$\Delta K = W > 0$

The kinetic energy starts to increase.

Block sliding on a frictionless surface and encountering a spring.

It touches the spring.

Compresses the spring, spring applies an opposite force on the block.

Velocity of the block decreases, because of spring.

Block of mass m slides on a surface with friction.

Because of work done due to friction, the kinetic energy of the block $K=\frac{1}{2} m v^2 = 0$

Certain type of forces, whose work can be stored as energy.

Potential energy, V

Work energy theorem: $W=\Delta K$

$\Delta K+\Delta V=0$

$W=-\Delta V$, work done by special forces or external forces.

Work done by force = – change in Potential energy $= -\Delta V$

If the system change from configuration 1 to configuration 2 because of a force being applied then the change in potential energy = – Work done by that forces.

$\Delta V=-W=-\int_{x_i}^{x_f} F d x$

$x_i$ is state 1, $x_f$ is state 2.

$\Delta V=-\int_{\substack{x_1 \ or \ x_i}}^{x_2 or x_f} F d x$

$x_i \text { or } x_f$ is the reference state 0.

$V(x)-V\left(x_0\right)=-\int_{x_0}^x F(x) d x$

Change in potential energy.

Reference value of $V\left(x_0\right)$ is not important.

If choose $x_0=0, V\left(x_0\right)=0$

Concept of potential energy is applicable only to those forces whose work is stored as energy.

$\rightarrow$ conservative forces

$V(x)-V\left(x_0\right)=-\int_{x_0}^x F d x$

Differentiate:

$\frac{d V}{d x}=-F(x)$

a) Work done by a conservative force depends only on initial and final position and not on the path taken.

Eg: work doNe by gravity, function of positions 1 and 2 and not the path.

If work done by a force depands on the path, then the force is not conservative and we cannot define a potential energy.

Dimension of V, same as work done or energy= $M L^2 T^{-2}$.

Then work done by conservative force as the body start from position A & comes back to some position (closed loop).

Body travels a path and comes back to its orignal position, if a conservative force action on the body.

$V_{f}- V_{i} = 0$

then work done by conservative force as the body start from position A & comes back to some position (closed loop)= 0

$W_{a-b}+W_{b-a}=0$

If f is corservative

$W_{a-b} = -W_{b-a}$

Body travels a path and comes back to its origmal position, if a conservative force action on the body

V = mgh

h is positive upwards (opposite to gravity).

$V_B-V_A=m g h$

$V_A=0$

$V_B=-m g h_1$

$V_B=V_A+m g \delta$

$\delta$ is vertical distance between B and A.

$-W=\Delta K$

$-\Delta V$

Calculate V due to gravity all we need is the vertical height of the body.

b) Spring which act applies a force

$F_{sp}=-kx$

Spring force is conservative, we can define a V associated with, $F_{sp}$.

x = 0 is the unstretched position of spring.

Work done by spring force = $\int_0^{x_m} F_s d x$

$-\int_0^{x_m} k x d x =-k\left[\frac{x_m^2}{2}-0\right]=-k \frac{x_m^2}{2}$

Work done by spring $=-\Delta V$

$\Delta V=k \frac{x_m^2}{2}$

If a spring is compressed by an amount $\delta$,

$V=\frac{1}{2} k \delta^2$

Even if the spring is extended by an amount $\delta$,

$V=\frac{1}{2} k \delta^2$

If spring (linear spring)

$F_{sp} = - kx$

$V=\frac{1}{2} k \delta^2$

$\delta$ is the displacement of the spring with respect to its unstretched length.

Potential energy due to force of gravity between two bodies.

(universal law of gravitation)

$F_{\text {gravity }}=-\frac{G_1 m_1 m_2}{r^2}$

$\Delta K=W$

Work done will be due to several forces.

Some of these forces will be conservative forces others will be non-conservative.

$\Delta k= W_{\text {cons. }}+W_{\text {non. cons. }}$

$W_{\text {cons. }} = -\Delta V$

$\Delta K+\Delta V=W_{\text {non. cons.}}$

If there are no non-conservative forces

$\Delta K+\Delta V=0$

Principle of conservation of mechanical energy.

$W_ {\text{non cons.}}=-\Delta U_{\text{internal energy}}$

$\Delta K+\Delta V+\Delta U=0$