Wg = -( PEgf - PEgi ) = -ΔPEg
Wg is work done by gravitational force in moving an object from a certain initial location to a certain final location. PEgi and PEgf are gravitational potential energies at the initial and final locations of the object respectively. An expression for gravitational potential energy in terms of position coordinate(s) may be obtained by obtaining an expression for the work done by gravitational force in terms of position coordinate(s).
Suppose an object of mass m is displaced from the location ( xi , yi ) to the location ( xf , yf ) under the influence of gravitational force. The work done by gravitational force is independent of the path followed from the initial to the final location because gravitation force is a conservative force. We may pick the easiest path to obtain an expression for the work done by gravitational force during this displacement. Let's take the vertical path that takes from the initial location ( xi , yi ) to ( xi , yf ) and then the horizontal path that takes to the final location ( xf , yf ). Gravitational force (weight) is directed vertically down. The horizontal path need not be considered because over that path gravitational force and displacement are perpendicular to each other and the work done is zero. For the vertical path, with out loss of generality, let's assume the final location to be below the initial location. In that case, gravitational force and the displacement are both directed vertically down and the angle between them is zero. The magnitude of the displacement is equal to the difference between the initial and final y coordinates, yi - yf. The magnitude of gravitational force is equal to the weight of the object which is equal to m|g|. Applying the definition for work, Wg = Fg d cos θ , the following expression can be obtained.
Wg = -m|g|( yf - yi )
Work done by gravity depends on the y-coordinates of the initial and final locations only. By comparing this equation with Wg = -( PEgf - PEgi ), the gravitational potential energy at an arbitrary point whose y-coordinate is y is defined as follows:
PEg = m|g|y
PEg is the gravitational potential energy of an object of mass m located at a point whose y-coordinate is y. Potential energy depends on the choice of reference point because position does. An object may have different potential energies for different coordinate systems. But, customarily, the reference point (origin) is put on the ground which makes the y-coordinate synonomous with the height of the object. The mechanical energy of an object under the influence of gravitational force ( MEg ) may now be expressed as
MEg = mv 2 ⁄ 2 + m|g|y
Example: An object of mass 10 kg is located on top of a 2 m table.
Calculate its gravitational potential energy with respect to the ground.
Solution: With respect to the ground means the reference point or the origin of the coordinate system should be fixed at the ground.
m = 10 kg ; y = 2 m ; PEg = ?
PEg = m|g|y = (10 * 9.8 * 2) J = 196 J
Calculate its gravitational potential energy with respect to the top of the table.
Solution: The origin of the coordinate system is fixed at the top of the table.
y = 0 m ; PEg = ?
PEg = m|g|y = (10 * 9.8 * 0) J = 0 J
Calculate its gravitational potential energy with respect to a point 5 m above the top of the table.
Solution: The origin is 5 m above the location of the object.
y = -5 m ; PEg = ?
PEg = m|g|y = (10 * 9.8 * -5) J = -490 J
Calculate the mechanical energy of the object with respect to the ground if it is sliding on the table with a speed of 2 m ⁄ s.
Solution: y = 2 m ; v = 2 m ⁄ s ; MEg = ?
MEg = mv 2 ⁄ 2 + m|g|y = ( 10 * 2 2 ⁄ 2 + 10 * 9.8 * 2 ) J = 216 J
Example: An object of mass 20 kg is dropped from a height of 50 m. Calculate the work done by gravitational force.
Solution: Let the origin of the coordinate system be fixed at the ground.
m = 20 kg ; yi = 50 m ; yf = 0 m ; Wg = ?
Wg = -m|g|(yf - yi ) = -20 * 9.8 * ( 0 - 50 ) J = 9800 J
Example: An object of mass 10 kg is displaced from the location (2, 4) m to the location (10, 34) m. Calculate the work done by gravitational force.
Solution: ( xi , yi ) = (2, 4) m. ( xf , yf ) = (10, 34) m.
m = 10 kg ; yi = 4 m ; yf = 34 m ; Wg = ?
Wg = -m|g|(yf - yi ) = -10 * 9.8 * (34 - 4) J = -29400 J
mvi2 ⁄ 2 + m|g|yi = mvf2 ⁄ 2 + m|g|yf
Example: An object is dropped from a height of 10 m. Use the principle of conservation of mechanical energy to calculate its speed by the time it hits the ground.
Solution: Let the origin of the coordinate system be fixed at the ground. Its initial speed is zero because it is released from rest.
yi = 10 m ; yf = 0 m ; vi = 0 m ⁄ s ; vf = ?
mvi2 ⁄ 2 + m|g|yi = mvf2 ⁄ 2 + m|g|yf
m|g|yi = mvf2 ⁄ 2
vf = √( 2|g|yi ) = √( 2 * 9.8 * 10 ) m ⁄ s = m ⁄ s
Example: An object is sliding down a frictionless 10 m 53° inclined plane. Calculate its speed by the time it reaches the ground.
Solution: The forces acting on the object are gravitational force and normal force exerted by the surface of the plane. The normal force is perpendicular to the displacement and thus the work done by the normal force does not contribute to the work done. Since the only force with non-zero contribution to the work done (gravity) is conservative, the principle of conservation of mechanical energy can be used.
Let the origin of the coordinate system be fixed at the final location (where it hits the ground) of the object. Then, ( xf , yf ) = ( 0, 0 ) m and ( xi , yi ) = ( 10 * cos 53°, 10 * sin 53° ) m = (6, 8) m. Its initial speed is zero because it is just sliding.
yi = 8 m ; yf = 0 m ; vi = 0 ; vf = ?
mvi2 ⁄ 2 + m|g|yi = mvf2 ⁄ 2 + m|g|yf
m|g|yi = mvf2 ⁄ 2
vf = √( 2|g|yi ) = √( 2 * 9.8 * 8 ) m ⁄ s = m ⁄ s
Example: An object is released from a frictionless roller coaster of height 30 m with an initial speed of 5 m ⁄ s. The roller coaster drops to the ground level and then rises to a height of 20 m and then falls to the ground level again. Calculate the speed of the object by the time it is at the top of the 20 m loop.
solution: The forces acting on the object are gravitational force and the normal force exerted by the roller coaster. The normal force does not contribute to the work done because it is perpendicular to the displacement. Only gravity contributes to the work done and hence the principle of conservation of mechanical energy can be used. Let the origin of the coordinate system be fixed at the ground.
yi = 30 m ; yf = 20 m ; vi = 5 m ⁄ s ; vf = ?
mvi2 ⁄ 2 + m|g|yi = mvf2 ⁄ 2 + m|g|yf
vf2 = vi2 + 2|g|(yi - y ) = {5 2 + 2 * 9.8 *(30 - 20)} m 2 ⁄ s 2 = 221 m 2 ⁄ s 2
vf = √(221) m ⁄ s = m ⁄ s
Ws = - (PEsf - PEsi ) = -ΔPEs
Ws is the work done by the force of a spring. PEsi and PEsf are the potential energies of the spring at the initial and final extensions (compressions) of the spring. The force due to a spring is related to the extension (compression) of the spring by a law known as Hook's Law. Hook's Law states that the force due to a spring is proportional but opposite in direction to the displacement (extension or compression) of the spring.
Fsx = -kx
Fsx is force exerted by a spring. x is the displacement of the spring as measured from the relaxed position of the spring. It is positive for extension and negative for compression. k is a proportionality constant for a given spring and is called Hook's constant of the spring. Its unit of measurement is N ⁄ m. k is always positive. This equation is a relationship between components. The force and the displacement can be positive or negative. The negative in the equation indicates that the force and the displacement have opposite signs or directions. If a relationship between the magnitudes is desired, the magnitudes of both sides should be equated.
Fs = k|x|
Example: A certain spring extends by 0.1 m when subjected to a 10 N force.
Calculate the Hook's constant of the spring.
Solution: Fs = 10 N ; |x| = 0.1 m ; k = ?
Fs = k|x|
k = Fs ⁄ |x| = (10 ⁄ 0.1) N ⁄ m = 100 N ⁄ m
What force would extend it by 0.35 m
Solution: |x| = 0.35 m ; Fs = ?
Fs = k|x| = ( 100 * 0.35 ) N = 35 m
An expression for the elastic potential energy of a spring may be obtained by first obtaining an expression for the work done by the force due to a spring. The force due to a spring varies as a function of the displacement. The formula for work done being used so far ( W = Fd cos θ ) can not be used in this case because it applies only if the force remains constant during the displacement. The work done by a variable force may be obtained from the graph of force versus displacement as the area enclosed between the force versus displacement curve and the displacement axis. Areas above the displacement axis are taken to be positive and areas below the displacement axis are taken to be negative. Since Fsx = -kx, the graph of force due to a spring versus displacement (extension) is a straight line in the fourth quadrant. The work done in extending a spring from its relaxed position to an arbitrary extension x is equal to the area of a triangle of base x and height kx and is negative which is equal to -kx 2. The work done in displacing a spring from an initial position xi to a final position xf may be obtained from the difference between the areas of the triangles associated with both triangles.
Ws = -(kxf 2 ⁄ 2 - kxi 2 ⁄ 2 )
Comparing this expression with the expression Ws = - (PEsf - PEsi ), the elastic potential energy ( PEs ) of a spring with arbitrary extension or compression x is defined as follows:
PEs = kx 2 ⁄ 2
The mechanical energy (MEs) of an object attached to a spring is given by
MEs = mv 2 ⁄ 2 + kx 2 ⁄ 2
Example: Calculate the elastic potential energy stored by a spring of Hook's constant 200 N ⁄ m when extended by 0.04 m.
Solution: k = 200 N ⁄ m ; x = 0.04 m ; PEs = ?
PEs = kx 2 ⁄ 2 = 200 * 0.04 2 ⁄ 2 J = 0.16 J
mvi 2 ⁄ 2 + kxi 2 ⁄ 2 = mvf 2 ⁄ 2 + kxf 2 ⁄ 2
Example: An object of mass 2 kg is attached to a spring of Hook's constant 100 N ⁄ m. The spring is compressed by 0.01 m and then released on a horizontal frictionless surface. Calculate the speed of the object when it is at the relaxed position of the spring.
Solution: m = 2 kg ; k = 100 N ⁄ m ; xi = -0.01 m ; xf = 0 m (relaxed position) ; vi = 0 m ⁄ s (released from rest) ; vf = ?
mvi 2 ⁄ 2 + kxi 2 ⁄ 2 = mvf 2 ⁄ 2 + kxf 2 ⁄ 2
kxi 2 = mvf 2
vf = √( k ⁄ m)|xi | = √( 100 ⁄ 2) * 0.01 m ⁄ s = m ⁄ s
Wnet = Wc + Wnc
Wnet is the net work done on the object. Wc and Wnc are the work dne by the conservative and non-conservative forces respectively. From the work kinetic energy theorem, net work done is equal to the change in the kinetic energy of the object and work done by a conservative force is equal to the negative of the change in potential energy. Therfore
Wnc = ΔKE + ΔPE = ΔME
Work done by a non-conservative force is equal to the change (loss) in the mechanical energy of the object. The effect of work done by a non-conservative force is to decrease the mechanical energy of the object. If gravity is the only conservative force acting on an object, then
Wnc = ( mvf 2 ⁄ 2 + m|g|yf ) - ( mvi 2 ⁄ 2 + m|g|yi )
And if force due to a spring is the only conservative force actitng on an object
Wnc = ( mvf 2 ⁄ 2 + kxf 2 ⁄ 2 ) - ( mvi 2 ⁄ 2 + kxi 2 ⁄ 2 )
Example: An object of mass 0.02 kg is dropped from a height of 20 m. By the time it hits the ground, its speed is 10 m ⁄ s. Calculate the work done by air resistance.
Solution: The forces acting on the object are gravity and air resistance. Air resistance is the non-conservative force. Let the coordinate system be fixed at the ground.
m = 0.02 kg ; yi = 20 m ; yi = 0 m ⁄ s ; vf = 10 m ⁄ s ; Wnc = ?
Wnc = ( mvf 2 ⁄ 2 + m|g|yf ) - ( mvi 2 ⁄ 2 + m|g|yi ) = {( 0.02 * 10 2 ⁄ 2 + 0 ) - (0 + 0.02 * 9.8 * 20)} J = -2.92 J
Wnc = ( mvf 2 ⁄ 2 + m|g|yf ) - ( mvi 2 ⁄ 2 + m|g|yi ) = { ( 0 + 0 ) - (10 4 * 20 2 ⁄ 2 + 0 ) } J = -2 * 10 6 J