Concept of work and the concept of kinetic energy.

Work done by a force.

If due to the action of this force, the body undergoes a displacement d.

One Dimensional case:

Define the work done as the product of the force F and the displacement d.

The particle moves a distance d in the same direction.

W = F d

Work done by Force F is the product of magnitude of Force and magnitude of displacement. (1D case)

We have a force $\vec{F}$ which is acting because of which particle undergoes a displacement $\vec{r}$.

1D case: $W=|\vec{F}||\vec{r}|$

$n$ forces can act on the body simultaneously.

Work done by $F_1$

Work done by $F_2$

Each force will do a work on the body.

$\phi \rightarrow$ The angle between $\overrightarrow{F}$ and $\vec{r}$.

W = Work done by $\vec{F}$

$W =\left|\overrightarrow{F}\right||\vec{r}| \cos \phi =\overrightarrow{F} \cdot \vec{r}$

$\vec{r}$= Displacement vector

$\text { a) } \phi=90^{\circ} \Rightarrow W=0$

$\text { b) } \phi>90^{\circ}, \quad W< 0$

Work done by a force $=\vec{F} \cdot \vec{r}$

Sometimes, we use $\vec{d}$ for displacement

$W=\vec{F} \cdot \vec{d}$

• $W \rightarrow$ Scalar, which can be either 0, positive, or negative.

• S.I unit of W = N.m = Joule

Example: A student lifts a rock of mass $10 \mathrm{kg}$ from the floor to raise it to a height of $2 \mathrm{m}$ (with negligible acceleration)

Identify:

a) Forces on the rock

b) Work done by each of these forces

Force by gravity = mg

It is moving up, the displacement is = 2 meters.

• $\phi=180^{\circ}$

Work done by gravity $=(\mathrm{mg})(2)(-1) =-196 \mathrm{J}$

Work done by $F$

$F=m g, d=2, \phi=0^{\circ}$

Force is acting in upward direction, displacement is in the same direction.

• $W=m g(2)=+196 \mathrm{J }$

$\Rightarrow$ Total work done on Rock by force external

$=196-196$ $=0 \mathrm{J}$

Total net external force on the rock was 0. So, therefore, the work done also has to be 0.

Force $F$ applied on a crate/ block of mass m. The coefficient of kinetic friction between block and ground $=\mu_k$

Block moves with a constant velocity and moves a distarce d.

a)Identify forces on the crate and their magnitude.

b) Find work done by each of these forces.

FBD of the block

Since $\vec{a} =0,$

• $\sum F_x =0$ and $\sum F_y =0$

mg = N

F = f

Since, there is relative motion between the block and the surface,

Friction force $f=\mu_k N$ $=\mu_k m g$

$F=f=\mu_k m g$

4 forces acting on block

a) mg acting downwards.

b) N =m g, normal reaction, acting upwards.

c) $F=\mu_k m g$, applied in the forward direction.

d) $f=\mu_k m g$, acting in the -x direction.

b) Find the work done by each of these four forces.

• $\vec{d} = \vec{r} = d\hat{i}$

$mg \downarrow N \uparrow$

'mg' and 'N' act in a direction perpendicular to the displacement d.

Work done by 'N' and 'mg' = 0

Force in $x$ direction

Work done by $F=F d \cos \left(0^{\circ}\right)=\mu_k m g d$

Work done by $f=f d \cos \left(180^{\circ}\right)=-\mu_k m gd$

Sum of all the four work done by different forces = 0.

Block of mass m is being pulled with a string on an incline.

Block moves a distance d.

Block being moved up a ramp on a frictionless surface with constant speed for a distance d.

a) Find force $F$ which cable must exert.

b) Work done by cable force $F$ so that the block moves a distance $d$.

c) Work done by gravity during this process.

d) If the block were to move vertically by the same height $h=d$ sin$\theta$, what would be work done by gravity.

Draw the free body diagram of the block.

Resolve mg along x and y.

$N=m g \cos \theta$

$F=m g \sin \theta$

a) Work done by

F = $m g \sin \theta d = m g \sin \theta d$

b) Work done by gravity:

$m g d \cos (90+\theta) = -m g d \sin \theta$

c) Resolve mg along x and y

$N=m g \cos \theta$

$F=m g \sin \theta$

d) If block were to be lifted vertically up by a distance h,

$\quad h=d \sin \theta,$

• $W_{gravity} =(m g)(d \sin \theta)(-1) = -m g d \sin \theta$

a) Draw FBD of particle, to identify forces acting on particle, and these forces will be along with their directions.

b) Find displacement vector $\vec{d}$ or $\vec{r}$ of the particle

c) Angle between $\vec{F}_i$ and $\vec{d}$ is $\phi_i$

Work by $F_i=\left|\vec{F}_i\right||\vec{d}| \cos \phi_i$

Sign of $W$

$W$ can be 0 positive or negative.

This definition of $W$ has been for assuming that force $\vec{F}$ is a constant.

We could have a case where $\vec{F}$ is not constant.

$\vec{F}$ function of displacement.

Generalize the concept of work done.

Force and displacement are along the same line $\longrightarrow$ ( $x$ axis)

Force $F$ depend on the displacement $x$.

Particle under the influence of $F$ moves from $x_i$ to $x_f$.

We divide region from $x_i$ to $x_f$.

We divide in small displacements $\Delta x$.

Let the average force on the region $\Delta x$ = $\bar{F(x)}$.

Work done in this small interal $\Delta x$ = $\bar{F}$ $(x) \Delta x$

$\delta W=\bar{F}(x) \Delta x$

Total work Done =$\Sigma \delta W$

Work done in this small interal $\Delta x$ = $\bar{F}$ $(x) \Delta x$

$\delta W=\bar{F}(x) \Delta x$

Total work Done =$\Sigma \delta W$ = $\sum \bar{F}(x) \Delta x$

In limit $\Delta x \rightarrow 0$, Summation becomes integral, $\sum \rightarrow \int$

$W=\int_{x_i}^{x_f} F(x) d x$

Work done is the area under the $F(x)$ v/s $x$ curve.

$x=0$ is at the uncompressed position of spring.

If we compress or stretch the spring, a force is applied by spring.

Spring force F(x) is a function of x.

Hook's law: $F(x)=-k x$

k is the spring constant, k>0

Force F is proportional to the extension or the compression of the spring,

F $\propto x$

F(x) =-k x

x >0 Extension

x <0 Compression

x =0 position of relaxed spring

SI unts of $k \rightarrow \mathrm{N} / \mathrm{m}$

If spring changs from $x_i$ to $x_f$ work done by spring froces.

W =$\int_{x_i}^{x_f} F_{s p} d x=\int_{x_i}^{x_4}-k x d x$

= -k $\int_{x_i}^{x_f} x d x=-\frac{k}{2} x^2|_{x_i}^{x_f}$=$\frac{1}{2} k\left(x_i^2-x_f^2\right)$

This is the work done by the spring.

Length of spring does not come into picture.

$\Rightarrow$ Work done by a force

=$|\vec{F}|| \vec{d} \mid \cos \phi$

$\vec{d}$ is displacement.

The work done depends on the reference frame in which we are observing the movement of the particle.

Kinetic energy is a property of a body, which a body gets by virtue of its speed.

If a body moves with a speed $v$,

Kinetic Energy $k \equiv \frac{1}{2} m v^2$.

where, $m$ is the mass of the body.

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

$\vec{v}$ = Velocity of particle

${v}\rightarrow$ Speed of particle

${v^2} =\vec{v} \cdot \vec{v}$

Unit of K = Same as unit of work done = J = $\left(\operatorname{kg} \frac{m^2}{s^2}\right)$

SI unit of kinetic energy is joules.

Kinetic energy being dependent on the speed is a frame dependent quantity. So, kinetic energy is also dependent on the reference frame.