UY1: Work-Energy Theorem

Kinetic Energy

Kinetic energy is the energy associated with the motion of a body.

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

Kinetic energy is a scalar and has the same units as work (J).

The work done by a force F in displacing a particle is equal to the change in the kinetic energy of the particle.

$$\begin{aligned} W &= K_{f}-K_{i} \\ &= \Delta K \end{aligned}$$

Derivation Of Equation For Kinetic Energy

Consider a block with mass $m$ and an initial velocity of $\vec{v}_{i}$. A force $F$ is applied on the block over a distance d. The final velocity of the block is $\vec{v}_{f}$.

$$\begin{aligned} d &= \frac{1}{2} \left( v_{i}+v_{f} \right) t \\ a &= \frac{v_{f}-v_{i}}{t} \end{aligned}$$

Let’s calculate the work done:

$$\begin{aligned} W &= Fd \\ &= \left( ma \right) d \\ &= m \left( \frac{v_{f}-v_{i}}{t} \right) \frac{1}{2} \left( v_{i} + v_{f} \right) t \\ &= \frac{1}{2} m v_{f}^{2}-\frac{1}{2}mv_{i}^{2} \end{aligned}$$

From the above, we can see that:

$$E_{k} = \frac{1}{2} mv^{2}$$


Power is the rate of doing work.

Average power:

$$P_{av} = \frac{\Delta W}{\Delta t}$$

Instantaneous Power:

$$\begin{aligned} P &= \lim_{\Delta t \rightarrow 0} \frac{\Delta W}{\Delta t} \\ &= \frac{dW}{dt} \\ &= \vec{F}.\frac{d \vec{s}}{dt} \\ &= \vec{F}.\vec{v} \end{aligned}$$

Units for power: watt, W = 1 $\text{J s}^{-1}$

1 horsepower = 746 W

Potential Energy

Potential energy is energy associated with the position (gravitational potential energy) or configuration (elastic potential energy) of an object.

Potential energy can be considered as stored energy that can be converted to kinetic energy or other forms of energy.

The potential energy concept can only be used when dealing with conservative forces.

Gravitational Potential Energy

Consider an object moving downwards from $y_{1}$ to $y_{2}$.

Work done by gravitational force is given by:

$$\begin{aligned}W_{\text{grav}} &= \vec{F}.\Delta \vec{y} \\ &= \left( -mg \right) \hat{j}. \left( y_{2}-y_{1} \right) \hat{j} \\ &= mgy_{1}-mgy_{2} \end{aligned}$$

If $m$ is moving upward from $y_{1}$ to $y_{2}$, the work done is given by the same formula.

We have:

$$U = mgy$$

We can write:

$$\begin{aligned} W_{\text{grav}} &= U_{1}-U_{2} \\ &=-\left( U_{2}-U_{1} \right) \\ &=-\Delta U \end{aligned}$$

If there are no other forces acting, then

$$\begin{aligned} W_{\text{tot}} &= W_{\text{grav}} \\ &=-\Delta U \\ &= U_{1}-U_{2} \end{aligned}$$

From work-energy theorem:

$$\begin{aligned} W_{\text{tot}} = \Delta K &=-\Delta U \\ K_{2}-K_{1} &= U_{1}-U_{2} \\ K_{1}+U_{1} &= K_{2}+U_{2} \end{aligned}$$

$$E = K + U = \text{constant}$$

If $m$ travels along a curved path, we can resolve along the usual $x-y$ direction to find the work done by the gravitational force.

$$\vec{f} = m \vec{g} =-mg \vec{j}$$

We resolve the distance moved into $\hat{i}$ and $\hat{j}$ components:

$$\Delta \vec{s} = \Delta x \, \hat{i} + \Delta y \, \hat{j}$$

The work done is then:

$$\begin{aligned} W &= \vec{f}.\Delta \vec{s} \\ &=-mg \, \hat{j} . \left( \Delta x \, \hat{i} + \Delta y \, \hat{j} \right) \\ &=-mg \Delta y \\ &=-mg \left( y_{2}-y_{1} \right) \\ &= mgy_{1}-mgy_{2} \\ &= U_{1}-U_{2} \end{aligned}$$

The above equations shows that:

  • Only the change in gravitational potential energy matters. (We are free to place the origin of coordinates or zero of potential energy in any convenient location.)
  • Work done by gravitational force is unaffected by any horizontal motion.
  • Only the positions of initial and final points in motion are important.
  • Work done by gravitational force depends only on its initial and final vertical coordinates (path independent).

Work Done On/By A Spring

From Hooke’s law:

$$\begin{aligned} F_{\text{spring}} &=-F_{\text{ext}} \\ &=-kx \end{aligned}$$

Work done on the spring (by external force) when the block is displaced from $x_{1}$ to $x_{2}$:

$$\begin{aligned} W &= \int\limits_{x_{1}}^{x_{2}} kx \, dx \\ &= \frac{1}{2} kx_{2}^{2}-\frac{1}{2}kx_{1}^{2} \end{aligned}$$

Work done by the spring when the block is displaced from $x_{1}$ to $x_{2}$:

$$W_{\text{elastic}} = \frac{1}{2} k x_{1}^{2}-\frac{1}{2} k x_{2}^{2}$$

Hence, for elastic potential energy, we have:

$$U = \frac{1}{2} k x^{2}$$

Note that:

  • There is no negative value
  • We cannot arbitrary assign zero value. $U = 0$ when $x = 0$ (spring is neither extended nor compressed)

Next: Potential Energy & Conservative Force

Previous: Concept Of Work

Back To Mechanics (UY1)

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