Unit 7: Kinetic Energy and Work

7.1 - Kinetic Energy

The kinetic energy KK associated with the motion of a particle of mass mm and speed vv, where vv is well below the speed of light, is:

K=12mv2K = \frac{1}{2}mv^2

The SI unit of kinetic energy is the joule (J), defined as:

1 joule=1J=1kgm2/s2\textrm{1 joule} = 1 \textrm{J} = 1 \textrm{kg} * \textrm{m}^2/\textrm{s}^2

7.2 - Work and Kinetic Energy

Work WW is energy transferred to or from an object via a force acting on the object. Energy transferred to the object is positive work, and from the object, negative work.

When two or more forces act on an object, their net work is the sum of the individual works done by the forces, which is also equal to the work that would be done on the object by the net force Fnet\vec F_{net} of those forces.

For a particle, a change ΔK\Delta K in the kinetic energy equals the net work WW done on the particle:

ΔK=KfKi=W\Delta K = K_f - K_i = W

In which KiK_i is the initial kinetic energy of the particle and KfK_f is the kinetic energy after the work is done. This is known as the work-kinetic energy theorem. This formula can also be rearranged as the following:

Kf=Ki+WK_f = K_i + W

The work done on a particle by a constant force F\vec F during displacement d\vec d is:

W=Fdcosϕ=FdW = Fd \cos \phi = \vec F \cdot \vec d

in which ϕ\phi is the constant angle between the directions of F\vec F and d\vec d.

Only the component of F\vec F that is along the displacement d\vec d can do work on the object.

7.3 - Work Done by the Gravitational Force

The work WgW_g done by the gravitational force FgF_g on a particle-like object of mass mm as the object moves through a displacement d\vec d is given by:

Wg=mgdcosϕW_g = mgd \cos \phi

in which ϕ\phi is the angle between Fg\vec F_g and d\vec d.

The work WaW_a done by an applied force as a particle-like object is either lifted or lowered is related to the work WgW_g done by the gravitational force and the change ΔK\Delta K in the object's kinetic energy by:

ΔK=KfKi=Wa+Wg\Delta K = K_f - K_i = W_a + W_g

If Kf=KiK_f = K_i, that is, the object's kinetic energy did not change, then the equation reduces to:

Wa=WgW_a = -W_g

or in other words, the applied force transfers as much energy to the object as the gravitational force transfers from it.

7.4 - Work Done by a Spring Force

The force Fs\vec F_s from a spring is:

Fs=kd\vec F_s = -k\vec d

where d\vec d is the displacement of the spring's free end from its position when the spring is in its relaxed state (not compressed or extended), and kk is the spring constant (a measure of the spring's stiffness). If an xx axis lies along the spring, with the origin at the location of the spring's free end when the spring is in its relaxed state, we can write:

Fx=kxF_x = -kx

A spring force is thus a variable force: It varies with the displacement of the spring's free end. (Yay calculus!)

If an object is attached to the spring's free end, the work WsW_s done on the object by the spring force when the object is moved from an initial position xix_i to a final position xfx_f is:

Ws=12kxi212kxf2W_s = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2

If xi=0x_i = 0 and xf=xx_f = x, the equation could be written as:

Ws=12kx2W_s = -\frac{1}{2}kx^2

Written as an integral makes this easier to work with:

Ws=xixfFx dxW_s = \int_{x_i}^{x_f} -F_x \space dx
Ws=xixfkx dxW_s = \int_{x_i}^{x_f} -kx \space dx

7.5 - Work Done by a General Variable Force

When the force F\vec F on a particle-like object depends on the position of the object, the work done by F\vec F on the object while the object moves from an initial position rir_i with coordinates (xix_i, yiy_i, ziz_i) to a final position rfr_f with coordinates (xfx_f, yfy_f, zfz_f) must be found by integrating the force.

If we assume that components only depend on their respective coordinates, then the work is:

W=xixfFx dx+yiyfFy dy+zizfFz dzW = \int_{x_i}^{x_f} F_x \space dx + \int_{y_i}^{y_f} F_y \space dy + \int_{z_i}^{z_f} F_z \space dz

If F\vec F only has an xx component, then this reduces to:

W=xixfF(x) dxW = \int_{x_i}^{x_f} F(x) \space dx

7.6 - Power

The power due to a force is the rate at which that force does work on an object. If the force does work WW during a time interval Δt\Delta t, the average power due to the force over that time interval is:

Pavg=WΔtP_{avg} = \frac{W}{\Delta t}

Instantaneous power is the instantaneous rate of doing work:

P=dWdtP = \frac{dW}{dt}

In other words, Power is the derivative of Work.

For a force F\vec F at an angle ϕ\phi to the direction of travel of the instantaneous velocity v\vec v, the instantaneous power is:

P=Fvcosϕ=FvP = Fv \cos \phi = \vec F \cdot \vec v