Mod 6

Module-6: Conservative & Non-conservative forces 6.1 Work Done by a Constant Force:

 F makes an angle Φ with the x axis and acts on a particle whose  displacement along the x axis is d . In this case we define the work W done by the force on the particle In Figure 1 a constant force

as the product of the component of the force along the line of motion by the distance d the body moves along that line. Then

W = [F cos Φ ]d ,

[1]

F φ

F cos φ Figure 1: A force makes the block undergo a displacement In the terminology of vector algebra we can write [1] as

 W = F.d ,

[2]

where the dot indicates a scalar [or dot] product. Work can be positive or negative. If the particle on which a force acts has a component of motion opposite to the direction of the force, work done by that force is negative. 6.2 Work Done by a Variable Force – One Dimensional Case: We consider first a force

 F that varies in magnitude only. Let the force be given as a function of

position F(x) and assume that the force acts in the x-direction. Suppose a body is moved along the xdirection by this force. What is the work done by this variable force in moving the body from x1 to x2? In Figure 2 we plot F versus x. We can write the total work done by F in displacing a body from x1 to x2 as x2

W12 = ∫ F ( x)dx , [3] x1

F(X)

X X1

X2

Figure 2: Work done by a Variable Force As an example, consider a spring attached to a wall. The work done by the applied force in stretching the spring so that its endpoint moves from x1 to x2 is

1

x2

x2

x1

x1

W12 = ∫ F ( x)dx = ∫ (kx)dx =

1 2 1 2 kx 2 − kx1 2 2

If we let x1=0 and x2=x, we obtain x

W = ∫ (kx)dx = 0

 F

y

 F a

1 2 kx , [4] 2

φ

 ∆r

φ

b ∆r

x

0 Figure 3: F and φ change along a path. 6.3 Work Done by a Variable Force – Two Dimensional Case:

 F acting on a particle may vary in direction as well as in magnitude, and the particle may  move along a curved path as shown in Figure 3. Figure 3 shows the value of angle φ between F and   ∆ r at two locations. We can find the work done on the particle during a displacement ∆ r from   [5] dW = F . ∆ r = F cos φ ∆ r ,  The work done by the variable force F as the particle moves, say, from a to b is found from The force

b   b Wab = ∫ F . d r = ∫ F cos φ dr , a

[6a]

a

We cannot evaluate this integral in until we are able to say how F and φ in Eq. [6a] vary from point to point along the path; both are functions of the x- and y-coordinates of the particle in Fig. 3.   We can obtain another equivalent expression for Eq. [6a] by expressing F and d r in terms of their         components. Thus F = i F x + j F y and d r = i dx + j dy ; so, F .d r = F x dx + F y dy . Then we can write [6a] as b

∫

W ab = [ Fx dx + F y dy ] ,

[6b]

a

As an example of a variable force consider the case of a simple pendulum and evaluate [6b]. Figure 4   shows the forces acting on the particle m. The applied force is F , T is the tension in the cord and  m g is the weight of the particle.  The work done as the mass m moves from φ = 0 to φ = φ0 under the action of the force F is

2

W=

φ =φ0

φ =φ0

φ =0

φ =0

∫

  F.d r =

∫ F cos φ d r ,

[7a]

T

Or

φ

x =( l −h ) tan φ0 , y =h

∫[ F dx + F

W =

x

y

y dy ] ,

x =0 , y =0

direction of dr

y

[7b]

φ

Let us evaluate [7b]. From Fig. [4b]

F

Fx = T sin φ and mg = T cos φ Eliminating T between these relations gives us

(b)

Fx = mg tan φ We also note from Fig. [4b] that

l

F y = 0 . Substituting these values for

φ

φ0

mg

Fx and F y into [7b], yields x =( l − h ) tan φ0 , y = h

W=

∫

dr

mg tan φ dx

φ dy dx

x =0, y =0

Now

from

Fig.[4a]

we

see

tan φ = dy / dx , or tan φ dx = dy . With

O

x (a)

the help of this, we can write the

Figure 4: Dynamics of a simple pendulum.

y =h

above equation as W =

h

∫ mgdy = mgh .

y =0

One can also compute the work done in displacing the particle along the arc with constant speed by applying a force that is always directed along the arc. Here it will be simpler to work with Eq. [7a] using the tangential force and taking dr = Rdφ. Now F = Fx2 + F y2 = Fx = mg tan φ and dr = Rdφ φ0

φ0

∫

∫

W = [mg tan φ ][cos φ ][ Rdφ ] = mgR sin φ dφ = mgR[1 − cos φ 0 ] = mgh 0

0

Where we have used the fact that R[1 − cos φ 0 ] = h . Notice that both these results are the same as the work that would be done in raising a mass m vertically through a height h. 6.4 Kinetic energy and the work - energy theorem: So far we have dealt with unaccelerated objects. In such cases the resultant force acting on the object  is not zero. Let us first consider the case of a constant resultant force F . Such a force acting on a  particle of mass m will produce an acceleration “ a ”. Let us choose the x-axis to be in the common

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  direction of F and a . Let vi and vf be the speed of the particle at t = 0 and t = t respectively. Then we can write a=

v f − vi

v f + vi

and x =

t

2

.t

Then the work done is W = Fx = ma x = m[

v f − vi t

][

v f + vi 2

]t =

1 1 m v 2f − m v i2 = K f − K i , [8a] 2 2

This equation shows that the work done by the resultant force acting on a particle is equal to the change in the kinetic energy of the particle. Let us now consider the case where the resultant force vary in magnitude but not in direction. Take the displacement to be in the direction of the force which is along the x-axis. Work done by the resultant force in displacing the particle from xi to xf is   W = F .d r =

∫

xf

∫ Fdx

xi

We can write acceleration a as a=

dv dv dx dv = . = v. dt dx dt dx

Hence xf

W=

xf

∫ Fdx = ∫

xi

xi

dv mv dx = dx

vf

1

∫ mvdv = 2 mv

vi

2 f

−

1 mv i2 = K f − K i = ∆K , [8b] 2

which is the work – kinetic energy theorem. It may be pointed out that the work – kinetic energy theorem holds whether the net force is constant or variable. Notice that when the speed of the particle is constant, there is no change in kinetic energy and the work done by the resultant force is zero. With uniform circular motion, for example, the speed of the particle is constant and the centripetal force does no work on the particle. If the kinetic energy of a particle decreases, the work done on it by the resultant force is negative. The displacement and the component of the resultant force along the line of motion are oppositely directed. The work done on the particle by the force is the negative of the work done by the particle on whatever produced the force. This is a consequence of Newton’s third law of motion. Hence Eq. [8] can be interpreted to say that the kinetic energy of a particle decreases by an amount just equal to the amount of work which the particle does. 6.5 The meaning of Kinetic Energy: Suppose a particle starts from rest. Then its initial kinetic energy Ki is zero. In this case, we can write Eq. [8] as W = K f −0 = K f So the kinetic energy of a particle is equal to the total work that was done to accelerate it from rest to its present speed. Consider the case of a hammer-head that is dropped from rest and its kinetic energy is used to drive some object into the ground. Here the kinetic energy of the hammer-head is used to do work on the object and drive it into the ground. This gives us another interpretation of kinetic energy: the kinetic

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energy of a particle is equal to the total work that particle can do in the process of being brought to rest.

6.6 Power: The power P supplied by a force is the rate at which the force does work. Consider a particle moving    with instantaneous velocity v . In a short time interval dt, the particle has displacement d s = v dt . The  work done by a force F acting on the particle during this time interval is    dW = F .d s = F .v dt The power delivered to the particle is then P=

  dW = F . v , [9] dt

The SI unit of power, one joule per second, is called a watt [W]. Consider a net force Fx acting on a particle in one dimension. The rate at which this force does work is P = Fx v x . Substituting Fx=max we have P = Fx v x = ma x v x ,

[10]

Or ax =

P , mv x

[11]

Thus for a constant power P, the acceleration varies inversely as the speed. If we write a x = dv x / dt in Eq.[10], then P = ma x v x = mv x

dv x d 1 dK = [ mv x2 ] = dt dt 2 dt

Assuming that P is constant, and integrating over some time interval, we get P∆ t = ∆K ,

[12]

So the time it takes an automobile or airplane at constant power to accelerate from one speed to another speed is proportional to the change in the kinetic energy. 6.7 The case of automotive power: The power requirements of a gasoline – powered automobile are an important and practical example of the concepts developed in this chapter. Two forces oppose the motion of an automobile: rolling friction and air resistance. Consider a car whose mass is 1251 kg. So its weight = 1251x 9.8 = 12,260 N. So the resisting force of rolling friction on a level road is Froll = µr η = [0.015][12260 N] = 180 N Here we have defined µr , the coefficient of rolling friction as the horizontal force needed for constant speed on a flat surface divided by the upward normal force exerted by the surface. Typical values of µr are 0.01 to 0.02 for rubber tyres on concrete. The force of rolling friction is nearly independent of car speed.

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The air resistance force Fair is approximately proportional to the square of the speed and can be expressed by the equation Fair =

1 CAρ v 2 2

Where A = the silhouette area of the car [seen from the front], ρ = density of the air [ about 1.2 kg/m 3 at sea level at ordinary temperatures], v = the speed of car, C = drag coefficient that depends on the shape of the moving body [ Typical values of C for cars range from 0.35 to 0.50]. For this car Fair = [1/2][0.38][1.77 m2][1.2 kg/m3][v2=[0.40 N. s2/m2]v2 Power needed for the car for constant speed v is P = Fforwardv = [Froll + Fair] = [180 N + (0.40 N. s2/m2)v2]v From this expression, we can calculate the following results:

v [m/s]

Froll [N]

Fair[N]

Fforward[N]

P[kW]

P[hp]

10

180

40

220

2.2

2.9

15

180

90

270

4.1

5.5

30

180

360

540

16

22

In a typical car engine, about 65% of the heat released from gasoline combustion is wasted in the cooling system and the exhaust. Another 20% or so is converted to work that does nothing to propel the car; this includes work done to oppose friction in the drive train and to run accessories such as the air conditioner and power steering. This leaves only about 15% of the energy to do work against the rolling friction and air resistance. Burning one litre of gasoline releases about 3.5x107 J of energy. The available energy per litre is then [0.15][3.5x107 J/L] = 5.3x106 J/L. Let us now look at the fuel consumption in the 15 m/s case. The power required is 4.1 kW = 4100 J/s. In one hour [3600 seconds] the total energy required is [4100 J/s][3600 s]=1.5x107 J and during this one hour the car travels a distance of [15 m/s][3600 s]=54 km. So the amount of fuel consumed in one hour 1. 5 x10 7 J 5. 3 x10 6 J / L

= 2. 8 L

Suppose the car can go from zero to 60 mi/h[27 m/s] in 6.1 s. The final kinetic energy is then 1 1 K = mv 2 = [1251 kg ][27 m / s ] 2 = 4. 6 x 10 5 J 2 2 The average additional power required for acceleration is Pav =

4. 6 x10 5 J = 75 kW = 100 hp 6.1 s

6.8 Conservative Forces: A force is conservative if the total work it does on a particle is zero when the particle moves around any closed path returning to its initial position. Another alternative definition of conservative force can be stated in the following way:

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The work done by a conservative force on a particle is independent of the path taken as the particle moves from one point to another. To illustrate the idea of a conservative force, let us consider the following case: When you ride a ski lift to the top of a hill of height h, the work done by the lift on you is “mgh” and that done by gravity is “–mgh”. When you ski down the hill to the bottom, the work done by gravity is “+mgh” independent of the shape of the hill. The total work done by gravity on you during the round trip is zero independent of the path you take. The force of gravity exerted by the earth on you is called a conservative force. Now consider you and the earth to be a two-particle system. When a ski lift raises you to the top of the hill, it does work “mgh” on the system. This work is stored as potential energy of the system. When you ski down the hill, this potential energy is converted to kinetic energy of motion. 6.9 Potential – Energy Functions: Since the work done by a conservative force on a particle does not depend on the path, it can depend only on the endpoints 1 and 2. We can use this property to define the potential – energy function U that is associated with a conservative force. Note that when the skier skis down the hill, the work done by gravity decreases the potential energy of the system. In general, we define the potential energy function such that the work done by a conservative force equals the decrease in the potential – energy function:

  W = ∫ F. d s = − ∆U

y

Or Path A

s2

  ∆U = U2 − U1 = − ∫ F. d s ,

2

[13a]

s1

For infinitesimal displacement, we have

  dU = −F. d s ,

1

[13b] o

6.10 Gravitational Potential Energy Near the Earth’s Surface:

Path B x

Figure 5: Two paths in space connecting the points 1 and 2. If the work done by a conservative force along path A from 1 to 2 is W, then the work done on the return trip along path B must be –W, because the round- trip work is zero.

We can calculate the potential - energy function associated with the gravitational force near the

  F = −mg j , we have       dU = − F . ds = −[mg j ].[dx i + dy j + dz k ] = + mgdy

surface of the earth from Equation [13b]. For the force

Integrating, we obtain

U = ∫ mg dy = mgy + U 0 U = U 0 + mgy ,

[14]

Where U0, the arbitrary constant of integration, is the value of the potential energy at y = 0. Since only a change in the potential energy is defined, actual value of U is not important. We are free to choose U to be zero at any convenient reference point. For example, if the gravitational potential energy of the earth – skier system is chosen to be zero when the skier is at the bottom of the hill, its value when the skier is at a height h above that level is mgh. Or we could choose the potential energy to be zero

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when the skier is at sea level, in which case its value at any other point would be mgy, where y is measured from sea level. 6.11 Potential Energy of Spring: Another example of a conservative force is that of a stretched spring. Suppose we pull a block attached to a spring from a position x = 0 [equilibrium] to x1. The spring does negative work because its force is opposite to that of the direction of motion. If we then release the block, the spring does positive work as it accelerates the block towards its initial position. The total work done by the spring when the block reaches its initial position is zero independent of how far we stretched the spring [assuming we did not stretch the spring so far that it was damaged]. The force exerted by the spring is therefore a conservative force. We can calculate the potential – energy function associated with this force from Equation [13b]:

  dU = − F . d s = − Fx dx = −[− kx]dx = kx dx Then

U = ∫ kx dx =

1 2 kx + U 0 2

Where U0 is the potential energy when x = 0, that is, when the spring is un-stretched. Choosing U0 to be zero gives

U=

1 2 kx , 2

[15]

When we pull the block from x = 0 to x1, we must exert an applied force Fapp= kx to balance the spring force. The work we do is x1

Wapp = ∫ kx dx = 0

1 2 kx1 2

This work is stored as potential energy in the spring – block system. 6.12 Non-conservative Forces: Not all forces are conservative. An example of a non-conservative force is kinetic friction. Suppose you push a box around some closed path on a rough table so that the box ends up at its original position. The force of kinetic friction is always opposite the direction of motion, so the work it does is always negative, and the total round-trip work it does cannot be zero. Another example of a nonconservative force is a force applied by human agent. The work that you do in pushing a box around a closed path on a rough table is not generally zero. It depends on how great a force you decide to exert on the box. Thus, neither the force you exert nor the force of kinetic friction is conservative, and no potential – energy function can be defined for either. In the macroscopic world, nonconservative forces are always present to some extent, the most common being frictional forces, which decrease the mechanical energy of a system. However, the decrease in mechanical energy is found to be equal to the increases in thermal energy produced by the frictional forces. Another type of nonconservative force is that involved in the deformations of objects. When you bend a coat hanger back and forth, you do work on the coat hanger, but the work you do does not appear as mechanical energy. Instead, the coat hanger becomes warm. The work done in deforming the hanger is dissipated as thermal energy. Similarly, when a ball of putty is dropped to the floor, it warms as it deforms on impact, and the original potential energy appears as

8

thermal energy. If thermal energy is added to mechanical energy, the total energy is conserved even when there are frictional forces or forces of deformation. A third type of nonconservative force is associated with chemical reactions. When we include systems in which chemical reactions take place, the sum of mechanical energy plus thermal energy is not conserved. For example, suppose that you begin running from rest. Originally you have no kinetic energy. When you begin to run, internal chemical energy in you muscle is conserved to kinetic energy of your body, and thermal energy is produced. It is possible to identify and measure the chemical energy that is used. In this case, the sum of mechanical, thermal, and chemical energy is conserved. Even when thermal energy and chemical energy are included, the total energy of the system does not always remain constant. The energy of a system can change because of some from radiation, such as sound waves or electromagnetic waves. However, the increase or decrease in the total energy of a system can always be accounted for by the appearance or disappearance of energy somewhere else. Let Esys be the total energy of a given system, Ein be the energy that enters the system, and Eout be the energy that leaves the system. The law of conservation of energy then states: E in − E out = ∆E sys ,

[a] Law of conservation of energy

Alternatively, The total energy of the universe is constant. Energy can be converted from one form to another, or transmitted from one region to another, but energy can never be created or destroyed. Law of conservation of energy The total energy E of many systems familiar from everyday life can be accounted for completely by mechanical energy Emech , thermal energy energy Etherm, and chemical energy Echem. To be comprehensive and include other possible forms of energy, such as electromagnetic or nuclear energy, we include Eother, and write generally E sys = E mech + E therm + E chem + E other , [b] 6.13 Potential Energy and Equilibrium in One Dimension: For a general conservative force in one dimension,

    F = Fx i and d s = dx i , Equation [10b] can be

written as

  dU = −F.d s = −Fx dx The force is therefore the negative derivative of the potential – energy function:

Fx = −

dU , dx

[16]

We can illustrate this general relation for a block-spring system by differentiating the function

U=

1 2 dU d 1 2 kx . We get Fx = − = [ kx ] = −kx 2 dx dx 2

9

Figure 6: Plot of the potential-energy function U versus x for an object on a spring. A minimum in a potential energy curve is a point of stable equilibrium.

Figure 6 shows a plot of

U=

1 2 kx versus x for a block and spring. The derivative of this function is 2

represented graphically as the slope of the line tangent to the curve. The force is thus equal to the negative slope of the curve. At x=0, the force

Fx = −

dU is zero and the block is in equilibrium. dx

A particle is in equilibrium if the net force acting on it is zero. Condition for equilibrium When x is positive in the figure, the slope is positive and the force is Fx is negative. When x is negative, the slope is negative and the force is Fx is positive. In either case, the force is in the direction that will accelerate the block towards lower potential energy. If the block is displaced slightly from x=0, the force is directed back towards x=0. The equilibrium at x=0 is thus stable equilibrium. In stable equilibrium, a small displacement results in a restoring force that accelerates the particle back towards its equilibrium position.

U

x

Figure 7: A particle at x=0 on this potential-energy curve will be in unstable equilibrium because a displacement in either direction results in a force d

Figure 7 shows a potential-energy curve with a maximum rather than a minimum at the equilibrium point x=0. Such a curve could represent the potential energy of a skier at the top of a hill. For this curve, when x is positive, the slope is negative and the force Fx is positive, and when x is negative, the slope is positive and the force F x is negative. Again, the force is in the direction that will accelerate the particle towards lower potential energy, but this time the force is away from the equilibrium position. The maximum at x=0 in Figure 7 is a point of unstable equilibrium. In unstable equilibrium, a small displacement results in a force that accelerates the particle away from its equilibrium position.

10

U

x

Figure 8: Neutral equilibrium. The force is zero at x=0 and at neighboring points, so displacement away from x=0 results in no force, and the system re

Figure 8 shows a potential-energy curve that is flat in the region near x=0. No force acts on a particle at x=0, and hence the particle is at equilibrium; furthermore, there will be no resulting force if the particle is displaced slightly in either direction. This is an example of neutral equilibrium. In neutral equilibrium, a small displacement results in zero force and the particle remains in equilibrium. Example 1: In a television tube, an electron is accelerated from rest to a kinetic energy of 2.5 keV over a distance of 80 cm. [The force that accelerates the electron is an electric force due to the electric field in the tube.] Find the force on the electron, assuming it to be constant and in the direction of motion. Solution: W=F∆x=Kf –Ki = Kf = 2.5 keV. Therefore, F=W/∆x =2.5x103 eVx1.6x10-19J.eV-1/ 0.80 m = 5.0x10-16N

Example 2: A force Fx varies with x as shown in the Figure 9. Find the work done by the force on a Fx, N particle as the particle moves from x=0 to x=6 m. Solution: We find the work done by the force by calculating the area under the Fx-versus-x curve:

6

W= A =A1 + A2 =[5 N][4 m} +[1/2][5 N][2 m] = 20 J + 5 J = 25 J.

3

5 4

2 1 1

2 3

4 5

6

X, m

Example 3: A truck of mass m is accelerated from rest at t=0 with constant power P along a level road. [a] Find the speed of the truck as a function of time. [b] Show that if x=0 at time t=0, the position function x[t] is given by x =

8P 3 / 2 t . 9m

Solution: We can write Eq.[11] as vdv =

Integrating this Eq., we get

P dt m

v2 P = t 2 m

⇒ v =[

2P 1 / 2 1 / 2 ] t m

11

Now v =

∫

x = dx = [

Integrating this, we get

dx dt

⇒

dx = vdt = [

2P 1 / 2 1/ 2 ] t dt m

2P 1 / 2 1 / 2 8P ] t dt = [ ]1 / 2 t 3 / 2 m 9m

∫

17.5 X(t) = (3/2)(2P/m)1/2t3/2

Distance, m

15.0 12.5

P = 800 W

10.0

P = 400 W

7.5

P = 200 W

5.0 2.5

Example 4: Each of the two jet engines on a Boeing 767 airliner develops a thrust [a forward force on 0 the airplane] of 197,000 N. When the airplane is flying at 250 m/s, what does each engine develop? 0 know 1P = Fv2= [1.97x10 3 5 N][250 4 5 = 4.93x107 W = [4.93x107 W/746 W] hp=66,000 hp. Solution: We m/s] Time, S

Example 5: The force between two atoms in a diatomic molecule can be represented approximately by the potential-energy function

a a u = u 0 [( )12 − 2( ) 6 ] x x where U0 and a are constants. [a] At what value of x is the potential energy zero? [b] Find the force F x. [c] At what value of x is the potential energy a minimum? Show that Umin= - U0.

Solution: [a]

The value of x at which U(x) is zero is obtained by setting u = 0 and this gives

a a a a u 0 [( )12 − 2( ) 6 ] = 0 , or, [( ) 12 − 2( ) 6 ] = 0 . x x x x Solving for x, we get

a a [ ] 6 = 2 ⇒ = 21 / 6 x x

⇒x=

a 1/ 6

2

[b] The force between the atoms can be found from

Fx = −

dU 12 U0 a 13 a , which gives Fx = [( ) − ( ) 7 ] dx a x x

[c] Set Fx equal to zero and solve for x. This will give x = a. [d] Set x = a

in

a a u = u 0 [( )12 − 2( ) 6 ] x x

to get U = - U0.

12

13

Problem Sheet of Module 6 1.

If the magnitude of the force of attraction between a particle of mass m1 and one of mass m2 is given by mm F = k 12 2 x where k is a constant and x is the distance between the particles. Find [a] the potential energy function and [b] the work required to increase the separation of the masses from x = x1 to x = x1+d.

2.

The magnitude of the force of attraction between the positively charged nucleus and the negatively charged electron in the hydrogen atom is given by

F =k

e2 r2

where e is the charge of the electron, k is a constant, and r is the separation between electron and nucleus. Assume that the nucleus is fixed. The electron initially moving in a circle of radius R 1 about the nucleus, jumps suddenly into a circular orbit of smaller radius R2.[a] Calculate the change in kinetic energy of the electron, using Newton’s second law. [b] Using the relation between force and potential energy, calculate the change in potential energy of the atom. [c] Show by how much the total energy of the atom has changed in this process. 3.

The so-called Yukawa potential r0 U 0 e − r r0 r gives a fairly accurate description of the interaction between nucleons[that is, neutrons and protons, the constituents of the nucleus]. The constants r0 = 1.5x10-15metres and U0 = 50 MeV.[a] Find the corresponding expression for the force of attraction. [b] To show the short range of this force, compute the ratio of the force at r = 2r0, 4r0, and 10r0 to the force at r = r0. U (r ) = −

4.

5.

Show that when friction is present in an otherwise conservative mechanical system, the rate at which mechanical energy is dissipated equals the frictional force f times the speed v at that instant, or d [ K + U ] = − fv dt The potential energy corresponding to a certain two-dimensional force field is given 1 U ( x , y ) = k ( x 2 + y 2 ) . (a) Derive Fx and Fy and describe the vector force at each point in terms of 2 its Cartesian coordinates x and y. (b) Derive Fr and Fθ and describe the vector force at each point in terms of the polar coordinates r and θ of the point. (c) Can you think of a physical model of such a force?

6.

A shopping cart full of groceries sitting at the top of a 2.0-m hill begins to roll until it hits a stump at the bottom of the hill. Upon impact, a 0.25-kg can of peaches flies horizontally out of the shopping cart and hits a parked car with an average force of 500 N. How deep a dent is made in the car (i.e., over what distance does the 500 N force act upon the can of peaches before bringing it to a stop)?

7.

Two industrial spies sliding an initially stationary 225 kg floor safe a distance of 8.50 m along a straight line toward their truck. The push F1 of spy-1 is 12.0 N, directed at an angle of 30 o downward from the horizontal; the pull F2 of spy-2 is 10.0 N, directed at 40o above the horizontal. The floor and safe make frictionless contact. (a) What is the total work done on the safe by forces F 1 and F2 during the 8.5 m displacement? (b) During the displacement, what is the work Wg done on the safe by its weight mg and what is the work WN done on the safe by the normal force N due to the floor? (c) The safe is initially stationary. What is its speed v at the end of the 8.50 m displacement?

8.

A 500 kg elevator cab is descending with speed vi = 4.0 m/s when the cube that controls it begins to slip, allowing it to fall with constant acceleration a = g/5. (a) During its fall through a distance d = 12m, what is the work w1 done on the cab by its weight mg? (b) During the 12m fall, what is the work W2 done on the cab by the upward pull T exerted by the elevator cable? (c) What is the total work W done on the cab in the 12 m fall? (d) What is the cab’s kinetic energy at the end of the 12 m fall?

9.

You apply a 4.9 N force to a block attached to the free end of a spring to keep the spring stretched from its relaxed length by 12 mm. (a) what is the spring constant of the spring? (b) What force does the spring exert on the block if you stretch the spring by 17 mm? (c) How much work does the spring force do on the block as the spring is stretched 17 mm as in (b)? (d) With the spring initially stretched by 17 mm, you allow the block to return to 0 mm (the spring returns to its relaxed state); you the

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compress the spring by 12 mm. How much work does the spring force do on the block during this total displacement of the block? 10.

You ski downhill on waxed skis that are nearly frictionless. (a) What work is done on you as you ski a distance S = 50m down the hill? (b) What is your speed on reaching the bottom of the run? Assume the length of the ski run is S = 50 m, its angle of incline is θ = 27o, and your mass is m = 50 kg. (The height of the hill is then h = S sinθ).

11.

Two blocks are attached to a light string that passes over a mass-less, frictionless pulley. The two blocks have masses m1= 3 kg and m2 = 5 kg and are initially at rest. Find the speed of either block when the heavier one falls a distance h = 4 m.

12.

A spring with a force constant k hangs vertically. A block of mass m is attached to the un-stretched spring and allowed to fall from rest. Find an expression for the maximum distance the block falls before it begins moving upward.

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