Work and Energy

Work is a measure of a force's ability to displace or deform an object. The work done by a force in displacing an object is defined to be the product of the component of the force in the direction of the displacement and the magnitude of the displacement.

W = F|| d

where W is the work done, F|| is the component of the force in the direction of the displacement and d is the magnitude of the displacement. If the angle between the force and the displacement is θ, then the component of the force in the direction of the displacement is equal to the product of the magnitude of the force and the cosine of the angle θ.

W = Fd cos θ

where F is the magnitude of the force. Work can be positive or negative depending on the value of θ. It is positive when θ is an acute angle and negative when θ is an obtuse angle. It is zero when the force is perpendicular to the displacement. The unit of measurement for work is the product of the unit of force (N) and unit of displacement (m) which is defined to be the Joule abbreviated as J.

Example: An object is pulled horizontally a distance of 5 m by a force of 100 N that makes an angle of 60° with the horizontal. Calculate the work done.

Solution: d = 5 m ; F = 100 N ; θ = 60° ; W = ?

W = Fd cos θ = (100 * 5 * cos 60°) J = 250 J

Example: An object is displaced horizontally a distance of 10 m while being pulled backwards by a force of 20 N that makes an angle of 37° with the horizontal. (An object can move in a direction that opposes the force if it has an initial velocity. The effect of the force is to slow down the object). Calculate the work done.

Solution: d = 10 m ; F = 20 N ; θ = (180 - 37)° = 143° ; W = ?

W = Fd cos θ = (20 * 10 * cos 143°) J = -160 J

Work Kinetic Energy Theorem

The net work done on an object is the sum of all the works done by all the forces acting on the object. In other words, it is the work done by the net force acting on the object.

Wnet = W1 + W2 + . . .

where Wnet is the net work done on the object and W1 , W2 , . . . are the works done by the forces F1 , F2 , . . . respectively. Or

Wnet = Fnet d cos θnet

where Fnet is the magnitude of the net force Fnet = F1 + F2 + . . . and θnet is the angle between the displacement and the net force.

Example: A 10 kg object is displaced by a distance of 5 m on a horizontal frictionless surface under the influence of the following forces: A 10 N horizontal force, A 20 N force that makes an angle of 60° with the horizontal and a 5 N force that is pulling backwards horizontally. Calculate the net work done on the object.

solution: The forces acting on the object are its weight, normal force and the forces listed. The weight and the normal force are perpendicular to the displacement and hence the work done by its weight and the normal force does not contribute to the net work done.

d = 5 m ; F1 = 10 N ; θ1 = 0° ; F2 = 20 N ; θ2 = 60° ; F3 = 5 N ; θ3 = 180° ; Wnet = W1 + W2 + W3 = ?

W1 = F1 d cos θ1 = (10 * 5 * cos 0°) J = 50 J

W2 = F2 d cos θ2 = (20 * 5 * cos 60) J = 50 J

W3 = F3 d cos θ3 = (5 * 5 * cos 180°) J = -25 J

Wnet = W1 + W2 + W3 = (50 + 50 - 25) J = 75 J

For simplicity let's assume that the net force acting on an object and its displacement are parallel.

Wnet = Fnet d

According to Newton's second Law Fnet = ma.

Wnet = mad

The product ad can be related to the initial and final speed of the object by using one of the equations of a uniformly accelerated motion.

vf 2 = vi 2 + 2ad

ad = ( vf 2- vi 2 ) ⁄ 2

Substituting for ad, the following expression for the net work done is obtained.

Wnet = mvf 2 ⁄ 2 - mvi 2 ⁄ 2

The expression mv2 ⁄ 2 is defined to be the kinetic energy of an object of mass m moving with a speed v.

KE = mv2 ⁄ 2

where KE stands for kinetic energy. The unit of measurement for energy is the joule (J). With this definition of kinetic energy, the net work done on an object is equal to the change of its kinetic energy.

Wnet = KEf - KEi = ΔKE = mvf 2 ⁄ 2 - mvi 2 ⁄ 2

This is a mathematical statement of the work kinetic energy theorem. The work kinetic energy theorem states that the net work done on an object is equal to the change of its kinetic energy.

Example: Calculate the kinetic energy of an 8 kg moving with a speed of 5 m ⁄ s.

Solution: m = 8 kg ; v = 5 m ⁄ s ; KE = ?

KE = mv2 ⁄ 2 = (8 * 52 ⁄ 2) J = 100 J

Example: The speed of a 4 kg object changed from 10 m ⁄ s to 5 m ⁄ s. Calculate the net work done on the object.

Solution: m = 4 kg ; vi = 10 m ⁄ s ; vf = 5 m ⁄ s ; Wnet = ?

Wnet = mvf 2 ⁄ 2 - mvi 2 ⁄ 2 = ( 4 * 52 ⁄ 2 - 4 * 102 ) J = -150 J

Example: A 20 kg object is displaced by a distance of 2 m on a horizontal frictionless surface under the influence of the following forces: A 50 N horizontal force and a 20 N force that makes an angle of 37° with the horizontal. If it started from rest

  1. calculate the net work done on the object.

    Solution: The forces acting on the object are its weight, the normal force exerted by the ground and the forces listed. Its weight and the normal force do not contribute to the net work done because they are perpendicular to the displacement.

    d = 2 m ; F1 = 50 N ; θ1 = 0° ; F2 = 20 N ; θ2 = 37° ; Wnet = W1 + W2 = ?

    W1 = F1d cos θ1 = ( 50 * 2 * cos 0° ) J = 100 J

    W2 = F2d cos θ2 = ( 20 * 2 * cos 37° ) J = 32 J

    Wnet = W1 + W2 = ( 100 + 32 ) J = 132 J

  2. calculate its speed at the end of the displacement.

    Solution: m = 20 kg ; vi = 0 m ⁄ s ; vf = ?

    Wnet = mvf 2 ⁄ 2 - mvi 2 ⁄ 2 = mvf 2 ⁄ 2

    vf 2 = 2Wnet ⁄ m = 2 * 132 ⁄ 20 m2 ⁄ s2 = 13.2 m2 ⁄ s2

Conservative Forces

Conservative forces are forces for which the work done is independent of the path followed. The work done depends only on a value on its initial location and final location that we call the potential energy of the object at the given location. Potential Energy of an object is the energy an object possesses just because of its location. For example an object separated from earth has potential energy just because of its location. If the object is released, this energy is converted to kinetic (motion) energy. The unit of measurement for potential energy is the Joule. The work done by a conservative force over a closed path is zero because the initial and final loctions are identical. Examples of conservative forces are gravitational force, electrical force, magnetic force and the force due to a spring.

The work done by a conservative force is equal to the negative of the change of the potential energy associated with the force. Potential energy depends only on position coordinates. But the way the potential energy depends on position coordinates is different for different kinds of forces.

Wc = -ΔPEc = -( PEcf - PEci )

Wc is work done by a certain conservative force. PEc is potential energy associated with the conservative force. PEci and PEcf are the potential energies at the initial and final locations respectively.

Example: An object is moved from a location where the potential energy is 100 J to a location where the potential energy is 200 J. Calculate the work done by the conservative force associated with the potential energy.

Solution: PEci = 100 J ; PEcf ; Wc = ?

Wc = -( PEcf - PEci ) = -(200 - 100) J = -100 J

Principle of Consevation of Mechanical Energy

The mechanical energy of an object is defined to be the sum of the kinetic energy and potential energy of the object.

ME = KE + PE

ME is the mechanical energy of an object whose kinetic and potential energies are KE and PE respectively. The expression for the potential energy varies from force to force. Also, sometimes there may be more than one conservative forces involved. On the other hand, the expression for kinetic energy is always mv2 ⁄ 2.

ME = mv2 ⁄ 2 + PE

Example: An object of mass 4 kg has a speed of 20 m ⁄ s when at a point where the object has a potential energy of 50 J. Calculate its mechanical energy.

Solution: m = 4 kg ; v = 20 m ⁄ s ; PE = 50 J ; ME = ?

ME = mv2 ⁄ 2 + PE =( 4 * 202 ⁄ 2 + 50 ) J = 850 J

If all the forces with a non-zero contribution to the net work done acting on an object are conservative, then the net work done on the object is equal to the work done by the conservative forces. According to the work kinetic energy, the net work done on an object is equal to the change of its kinetic energy. And the work done by conservative forces is equal to the negative of the change in itspotential energy. Therefore, if all the forces acting on an object are conservative the following equation holds:

ΔKE = -ΔPE

But ΔKE = KEf - KEi and ΔPE = PEf - PEi . After substituting these expressions and collecting the initials on one side and the finals on the other side, the following equation is obtained.

KEi + PEi = KEf + PEf

The left hand side of this equation is equal to the initial mechanical energy of the object and the right hand side of thisequation is equal to the final mechanical energy of the object. The mechanical energy of the object remains the same. In other words the mechanical energy of the object is conserved. This is a mathematical statement of the principle of conservation of mechanical energy. The principle of conservation of mechanical energy states that if all the forces with none-zero contribution to the the work done acting on an object are conservative, then the mechanical energy of the object is conserved.

Example: A 5 kg object was displaced from a location where its potential energy is 20 J to a location where its potential energy is 10 J under the influence of conservative forces only. If its initial speed is 2 m ⁄ s,

  1. Calculate its initial kinetic energy.

    Solution: m = 5 kg ; vi = 2 m ⁄ s ; KEi = ?

    KEi = mvi 2 ⁄ 2 = ( 5 * 22 ⁄ 2 ) J = 10 J

  2. calculate its initial mechanical energy.

    Solution: PEi = 10 J ; MEi = ?

    MEi = KEi + PEi = ( 10 + 20 ) J = 30 J

  3. Calculate its mechanical energy at its final location.

    Solution: Since all the forces acting on the object are conservative, mechanical energy is conseved.

    MEf = ?

    MEf = MEi = 30 J

  4. calculate its kinetic energy at the end of the displacement.

    Solution: PEf = 10 J ; KEf = ?

    MEf = KEf + PEf

    KEf = MEf - PEf = ( 30 - 10 ) J = 20 J

  5. calculate its speed at the end of the displacement.

    Solution: vf = ?

    KEf = mvf 2 ⁄ 2

    vf = √( 2 KEf ⁄ m ) = √( 2 * 20 ⁄ 5) J = m ⁄ s