Chapter 6 Work and energy The work W done done by an agent exerting a constant force is defined as the product of the component of the force along the direction of displacement and the magnitude of the displacement. = ( F W = F cos θ) s = F.s
(6.1)
where θ is the angle between the force F force F and and the displacement s displacement s.. Work is a scalar quantity Note that work is a scalar quantity that has no direction. In fact, all types of energy are scalar quantities. This is one of the nice features of the energy approach. It is easier to deal with scalar quantities than with vector quantities! Zero work The work is zero if the force is zero , the displacement is zero, or the angle between F and s is 90o
When an object is displaced on a frictionless, horizontal surface, the normal force n and the gravitational force m force mg do no work on the object. In the situation shown in fig, F fig, F is the only force doing work on the object. cosθ θ is Negative work The work is negative if the force or one of its components is opposite to the displacement In the figure below the displacement is to the right, therefore (a) Represents no work is done, (b) most negative work, (c) most positive work (d) negative work
Note that work is energy transfer. transfer. • If W is positive energy e nergy is transferred to the system and the stored energy of the system increases; • if W is negative energy is transferred from the system and the stored energy of the system decreases
Example1 A vacuum cleaner being pulled at an angle of 30.0° from the horizontal by a force of 50N (a) calculate the work done as the vacuum cleaner moves a distance of 3.0m (b)What is the work done by the weight mg and the normal force n
Solution (a) F=50N s= 3m q= 30.0° W= Fs cos q = (50N)(3m )(cos 30.0°) 130 J o
(b) Since the angle between n and s is 90 and cos 90 is zero then the work is zero. o since the angle between mg and s is 270 , o and since cos 270 = 0, then the work done is zero.
Work done by a Varying Force When the force acting on an object is varying force i.e it depends on the position of the object the work done by this force while the object moves from an initial position x1 to a final position x 2 is given by W =
x 2
x 1
F x dx
(6.2)
Where F x is the component of the force in x direction i.e parallel to motion The work done by the force component F x for the small displacement x is F x x, which equals the area under the force displacement graph. In the given figure it is the area of the shaded rectangle. Figure (a) shows the total work done, for t he displacement from xi to x f is approximately equal to the sum of the areas of all the rectangles. In figure (b) the work done by the component F x of the varying force as the particle moves from xi to x f is exactly equal to the area under this curve.
Example-3
The force acting on a particle is constant for the first 4.0 m of motion and then decreases linearly with x from xB = 4.0 m to xC = 6.0 m. as shown in figure. Find the net work done by these forces as the object moves from A to C. Solution The net work done by these forces is the area under the curve. WAB= 5x4= 20 J WBC = ½ ( 2) ( 5)=5J WAC = 20 + 5 = 25J
The work done by a spring
When a body attached to a spring is displaced a distance x from its e quilibrium position, the spring will exert a force F on the body given by Hook’s law as: F = -k x
(6.3)
where k is a constant called the elastic force constant of the spring. The negative sign in equation 6.3 means that F and x are in opposite directions. F is called restoring force because it tends to restore the object to its equilibrium position. The work done by the spring on the displaced mass x
1
W = k x dx = 2 k x 2
(6.4)
0
Example 6.3 : A block lying on a smooth, horizontal surface is connected to a with force constant 80 N/m the is compressed a distance of 3.0 cm the equilibrium position. Calculate work done by the spring as the moves from xi =-3 cm , to xf = 0. Solution: -2
Xi = -3 cm = -3 x 10 m , Xf = 0 K = 8 N/m
,
2
Ws = ½ k x m -2 2 W = ½ (80) (-3 x 10 ) = 3.6 x -2 5 10 Joule = 3.6 x 10 ergs. The determination of the elastic force constant of a spring The elastic force constant of a spring is determined experimentally as follows 1. A mass m is suspended to a spring vertically. 2. Due to the weight of the mass the spring will be elongated by a distance d. Then the restoring force Fs = -kd 3. At equilibrium or balance. kd = mg ,
k = mg / d (6.5)
spring spring from the block
Example 6.4 : A mass of 0.55 kg stretches a spring a distance of 2.0 cm. What is the work done the spring is placed on a horizontal plane and compressed a distance of 4 cm. Solution: 2
k = mg / d = 0.55 x 9.8 /.02 = 2.7 x 10 N/m 2 2 -2 2 W = 1/2 k x m = 1/2 x 2.7 x 10 x (4 x 10 ) = 0.216 J. 6.5 Work and kinetic energy Consider a particle of mass m moving in x-direction with a constant force F acting on it. From Newton’s second law we have F = ma then , if the particle is displaced from xi = 0 , to xf = x the work done is given by :
W = Fx = m a x The acceleration is given by
a =
v − vo
where, v : is the final velocity , and v o : is
t
the initial velocity, and t is the time. The displacement x is given by x =
v + vo t
W = m (
=
1 2
mv
2
, therefore the work done is given by
v − vo
−
t 1
2
)(
v + vo
mv o
2
) t
2
The kinetic energy The product of half the mass and the square of the velocity is called the kinetic energy of the body. 2 K = ½ m v . It is a scalar quantity that has unit of Joule, which is the unit of work. The work energy theorem This theory states that: the work done on an object by the resultant force (sum of all forces ) acting upon it, equal the change in its kinetic energy . The work-energy theorem is valid for constant force as well as for varying forces. Note that: According to the work-energy theorem 1- work done by all forces = ∆K 2- if the work is positive K f > Ki and the velocity increases, v f > vi 3- if the work is negative Kf < Ki and the velocity decreases v f > vi 4- only when external force acts on an object its energy increases by an amount equal to the work done by this force. 5- we can consider the kinetic energy of a body as the work that the body do in coming to rest.
Power Power is defined as the rate of doing work or the rate of energy transfer. If external force acts on a system so that it does work ∆w in a time ∆t, the average power is given by:
p =
∆w ∆t
The instantaneous power is given by:
p = lim ∆ t → 0
∆w ∆ t
=
dw dt
For a constant force w = F.S
p =
d ( F . S ) d t
= F
d S d t
= F .v
Where v is the velocity. The units of power are 1- Joule/second = Watt. 2- k watt = 1000 watt, 3- hp = 746 watt. A unit of work and energy is kwh (kilowatt- hour).
Example What is the energy consumed by 100 watt bulb in one hour. solution p = w/t 5 w= pt = 100 x 3600 =3.6 x10 Joules.
Potential energy. The potential energy is the energy associated with the position of an object. It can be considered as a stored energy of the object t hat can be converted into work or kinetic energy.
The concept of potential energy can be used only when the object is acted by a type of force called conservative force . The gravitational force , the electric force and the force of a spring are all conservative force. When only external conservative forces act on an object the kinetic energy gained (or lost) equal the potential energy loss (or gain). This is known as the law of conservation of mechanical energy. Conservative force
A force is conservative if the work done by this force when acting on a moving object between two points is independent of the pat h the particle takes between these two points. The work done by conservative force depends only on the coordinates of the initial and final points and independent of the path taken between these two points. (see fig 6.
1
a
b
2 Fig 6.2
Let W ab (1) be the work done in moving from point a to point b through the path (1), and W ab (2) be the work done in moving from point a to point b through the path (2). If the force is conservative then W ab (1) = W ab (2) The work done by conservative force in a reversed path equal the negative of the work done by this force in the non-reversed path. W ab (1) = - W ba (1) = - W ba (2) W ab (1) = - W ba (2) W ab (1)+ W ba (2) =0 Since the path aba is a closed path then, we can say that: the work done by conservative force when the object moves in a closed path is zero.
The work done by gravitational force
1- When a body of mass m moves from a point of height hi to a point of height h f from the earth surface, under the effect of gravity, the work done is given by W = - mg ( h f - hi ) It is negative because the motion is against the weight mg. 2- If the body makes a round trip and returns to its initial height, h f = hi, then the work done by gravity is zero. Since the work done by any force equal the change in kinetic energy, then in this case the change in kinetic energy is zero. The final kinetic energy equal the initial kinetic energy and the final speed equal the initial speed. K f = K i v f = vi The work done by a spring
The work done by a spring to move a body from initial position xi to a final position x f is given by 1
1
W = 2 k x f 2 − 2 k xi2 If the body returns to its initial position then, xi = x f ,and W = 0. Nonconservative force A force is nonconservative if the work done by this force on a particle moving between two points depends on the path taken e.g. friction forces. The property of nonconservative force is that W ab (1) ≠ (2) ≠W ab W ab (1) + W ab (2) ≠ ≠ 0 Potential energy and the work done by conservative force The work done by the conservative force equal the decrease in potential energy. If the work is Wc, then
Wc = - ∆U i.e. the work done by the conservative force equal the negative of the change in potential energy. Conservation of mechanical energy When a conservative force acts on a moving object, t hen from the work energy theorem Wc = ∆K Where, Wc is the work done ∆K is the change in kinetic energy. Since the force is conservative then
Wc = - ∆U ∆U is the change in potential energy. From these two equations we get. ∆K = - ∆U ∆K + ∆U =0 ∆(K + U ) = 0 K + U = E is the total energy. If Ki is the initial kinetic enrgy. Ui is the initial potential enrgy. Kf is the final kinetic enrgy. Uf is the final potential enrgy K i+ U i = K f + U f Ei =E f The law of conservation of mechanical energy This law states that the total mechanical energy of a system remains constant if the only force acting on the body is conservative force. If the potential energy decreases (or increases ) the kinetic energy increases ( or decreases) by the same amount. Non-conservative force and the work energy theorem. In physical systems neoconservative forces such as friction, are usually present. So consider a case in which conservative and neoconservative forces acts on a system. Let W nc be the work done by the neoconservative force and Wc the work done by conservative force. The total work done by the two types of forces is given by W = W nc + W c From the work- energy theorem W =∆K, W nc + W c= ∆ k But, the work done by the conservative force e quals negative the change in potential energy then W nc -∆U= ∆ k ∆( U + k) = E f -Ei W nc = ∆U + ∆ k = ∆( From the last equation we see that, the work done by all nonconservative forces equal the change in the total mechanical energy of the system. Note that if no nonconservative force acts on the system W =0 then E f =E .i That is the total mechanical energy is conserved .
Problems 6.1 If a man lifts a 20 kg bucket from a well and does 6 kJ of work, how deep is the well? assume the speed of the bucket remains constant as it is lifted. 6.2 A 65 kg woman climbs a flight of 20 stairs, each 23 cm high. How much work was done against the force of gravity in the process? 6.3 A tugboat exerts a constant force of 500 N on a ship moving at constant speed through a harbor. How much work does the tugboat do on yhe ship in a distance of 3 km? 6.4 Verify the following energy unit conversions: 7 (a) 1J = 10 ergs. (b) 1J= 0.737 ft.lb. 6.5 A cheerleader lifts his 50 kg partner straight up off the ground a distance of 0.6 m before releasing her. If he does this 20 t imes, how much work has he done on her? 6.6 A 100 kg slid is dragged by a team of dogs a distance of 2km over a horizontal surface at a constant velocity. If the coefficient of friction between the slid and the snow is 0.15, find the work done by:(a) The team of dogs (b) The force of friction. 6.7 A horizontal force of 150 N is used to push a 40 kg box on a rough, horizontal surface through a distance of 6 m. If the box moves at constant speed, find (a) the work done by the150 N force, (b) the work done by friction, and (c) the coefficient of kinetic friction. 6.8 A 15 kg block is dragged over a rough, horizontal surface by a constant force of o 70 N acting at an angle of 20 above the horizontal. The block is displaced a distance of 5 m, and the coefficient of kinetic friction is 0.3. Find the work done by (a) 70 N force, (b) the force of friction (c) the normal force, and (d) the force of gravity. (e) what is the net work done on the block? 6.9 If you push a 40 kg crate at a constant speed of 1.4 m/s across a horizontal floor (µk = 0.25), at what rate is work being done on the crate by (a) you? (b) the frictional force? 6.10 Batman, whose mass is 80 kg, is holding onto the free end of a 12 m rope, the other end of which is fixed to a tree limb above. He is able to get the rope in motion as only batman knows how, eventually getting it to swing enough that he can reach a o ledge when the rope makes a 60 angle with the downward vertical. How much work was done against the force of gravity in this maneuver? 6.11 A cart loaded with bricks has a total mass of 18 kg and is pulled at constant o speed by a rope. The rope is inclined at 20 above the horizontal and the cart moves on a horizontal plane. The coefficient of kinetic friction between the ground and the cart is 0.5. (a) What is the tension in the rope? when the carts is moved 20 m, (b) How much work is done in the cart by the rope? (c) How much work is done by the friction force?
6.12 A 0.6 kg particle has a speed of 2m/s at a point A and kinetic energy of 7.5 J at point B. What is (a) its kinetic energy at point A? (b) its velocity at point B? (c) the total work done on the particle as it moves from A to B? 6.12 Calculate the kinetic energy of a 100 kg satellite orbiting the earth at a speed of 7000 m/s? 6.13 A 40 kg box initially at rest is pushed a distance of 5m along a rough horizontal floor with a constant applied horizontal force of 130 N. If the coefficient of friction between the box and floor is 0.3, find (a) the work done by the applied force, (b) the work done by friction (c) the change in kinetic energy of the box, and the final speed of the box? o
6.14 A block of mass 0.6 kg slides 6.0 m down a frictionless ramp inclined at 20 to the horizontal it then travels on a rough horizontal surface where k = 0.5. (a) What is the speed of the block at the end of the incline (b) What is its speed after traveling a distance of 1.0 m on the rough plane. (c) What distance does it travel on this horizontal plane before coming to rest?
