CHAPTER 3 FORCES & NEWTON S LAW OF MOTION

CHAPTER 3 FORCES & NEWTON S LAW OF MOTION 3. The concepts o orce and mass FORCE A orce is a push or pull upon an object resulting rom the object's interaction with another object. Force is a quantity which is measured using the standard metric unit known as the Newton, N. Force is a vector quantity. There must be a net orce (unbalanced orce) acting on an object or the object to change its velocity Net orce, F, is the vector sum or the unbalanced or acting on an object. Force can be divided into: a. Contact orces-physical contact between objects-rictional orces, tensional orces, normal orces b. Non-contact orces @ action -at-a-distance orces two objects are not in physical contact with each other-gravitational orces Examples o action-at-a-distance orces include gravitational orces (e.g., the sun and planets exert a gravitational pull on each other despite their large spatial separation; even when your eet leave the earth and you are no longer in contact with the earth, there is a gravitational pull between you and the Earth), electric orces (e.g., the protons in the nucleus o an atom and the electrons outside the nucleus experience an electrical pull towards each other despite their small spatial separation), and magnetic orces (e.g., two magnets can exert a magnetic pull on each other even when separated by a distance o a ew centimeters. MASS

A measure o an object s inertia. Quantity o matter The dierence between mass & weight are show below: Mass Unit : kg Not reerred to gravity Weight Unit : Newton Always reerred to gravity ( Giancoli 6 th Ed. Pg 75) EXAMPLE 3.: A book on a horizontal desk. Free body diagram There are two orces acting on the book: i. Gravity pulls downwards so the weight, W is always present, W = mg ii. Table pushes upwards and a orce result is normal orce, N which is perpendicular to the contact surace. 3. Newton s Laws o Motion Newton s irst law I no orces act on an object, it continues in its original state o motion; that is, unless something exerts an external orce on it, an object at rest remains at rest and an object moving with some velocity continues with that same velocity. Newton s irst law also known as the Law o Inertia. Inertia is a property o a body to make the body to remain in its state, either stationary or moving with constant velocity. Mass is a measure o inertia. It is a physical quantity that determines how diicult it is to accelerate or decelerate an object. SI unit is kg. Newton s second law When a net external orce F acts on an object o mass m, the acceleration a that results is directly proportional to the net orce and inversely proportional to the mass. The direction o the acceleration is the same as the direction o the net orce. F a _ or F ma m SI unit o orce: kg.m/s = Newton, N

Newton s third law: When two bodies interact, they exert equal but opposite orces on each other. Example: Forces come in pairs, i a hammer exerts a orce on a nail; the nail exerts an equal but oppositely directed orce on the hammer. i.e. F AB = F BA Action orce = Reaction orce Note that they act on dierent bodies, so they will not cancel. 3.3 Types o orces The Gravitational Force The gravitational orce is a mutual orce o attraction between any two objects. Expressed by Newton s Law o Universal Gravitation: For two particles that have mases m and m are sparated by a distance r, the orce that each exerts on the other is directed along the line joining the particles and has a magnitude given by mm F G where G is the universal gravitational constant r G = 6.673 x 0 - Nm /kg The weight o an object on or above the earth is the gravitational orce that the earth exerts on the object: M Em W G where M E is the mass o the earth r Gravity Near the Earth s Surace M Em M E g = W G r The Normal Force The normal orce N, is the orce that a surace exerts on an object with which it is in contact. The normal orce is exerted perpendicular to the surace. The Tension Force Tension T, is the orce which is transmitted through a string, rope, or wire when it is pulled tight by orces acting at each end. The tensional orce is directed along the wire and pulls equally on the objects on either end o the wire. 3

The Frictional Force When an object is in motion on a surace or through a viscous medium, there will be a resistance to the motion. The resistance is called the orce o riction. The riction orce is the orce exerted by a surace as an object moves across it or makes an eort to move across it. The riction orce opposes the motion o the object. For example, i a book moves across the surace o a desk, the desk exerts a riction orce in the direction. Static Friction, ƒs Static riction acts to keep the object rom moving. The magnitude o the static riction depends on the magnitude o the applied orce and can have the values up to a maximum o s us N where u s = the coeicient o static riction Kinetic Friction, ƒ k The orce o kinetic riction acts when the object is in motion. The magnitude given by k uk N where u k = the coeicient o kinetic riction EXAMPLE 3. A orce o 48.0N is required to start a 5.0kg box moving across a horizontal concrete loor. A) What is the coeicient o static riction between the box and the loor? B) I the 48.0N orce continues, the box accelerates at 0.7 m/s. What is the coeicient o kinetic riction? Solutions: A ree-body diagram or the box is shown. Since the box does not accelerate vertically, F mg N (a) To start the box moving, the pulling orce must just overcome the orce o static riction, and that means the orce o static riction will reach its maximum value o F F. Thus we have or r s N the starting motion, F 48.0 N P F F F mg 0.98 P r s N s s mg 5.0 kg 9.8m s (b) The same orce diagram applies, but now the riction is kinetic riction, and the pulling orce is NOT equal to the rictional orce, since the box is accelerating to the right. F F F ma F F ma F mg ma F P r P k N P ma 5.0 kg9.8m s 48.0 N 5.0 kg 0.70 m s P k mg 0.9 k Fr F N FP mg 4

EXAMPLE 3.5: Calculate the acceleration due to gravity on the Moon. The Moon s radius is.74 0 6 m and its mass is 7.35 0 kg. Solutions: The orce o gravity on an object at the surace o a planet is given by Newton s law o Universal Gravitation, using the mass and radius o the planet. I that is the only orce on an object, then the acceleration o a reely-alling object is acceleration due to gravity. M m Moon F G mg G Moon r Moon 7.35 0 kg 6.74 0 m M g G 6.67 0 N m kg.6 m s r Moon Moon Moon EXAMPLE 3.6: Two blocks are in contact on a rictionless table. A horizontal orce is applied to one block, as shown in igure. (a) I m.3 kg, m. kg, and F 3. N, ind the orce o contact between the two blocks. (b) Show that i the same orce F is applied to m, but not to m, the orce o contact between the blocks is. N, which is not the same value derived in (a). Explain. Answer: F m N N m a) Consider mass m F N ma () Consider mass m m a () N () + () gives F m m a F a m m m F m m N (3) 5

or m.3 kg, m. kg, and F 3. N, N.0 N b) Interchange m and m, one obtains N.N In order to preserve constant acceleration, greater orce o the contact is needed to move the mass m. EXAMPLE 3.7: The two blocks, m = 6 kg and M = 88 kg, shown in igure are ree to move. The coeicient o static riction between the blocks is s 0. 38, but the surace beneath M is riction-less. What is the minimum horizontal orce F required to hold m against M? Answer: The ree-body diagrams or both m and M are shown to the right. When N s s, max s, F F where N is the normal orce exerted by m on M. The equations o motion or blocks m and M are Fmin s mg N ma and N Ma respectively. The vector equation above or block m in component orm is m Fmin N ma N F M s s N mg; while the equation or M can be written as mg N Ma. Ater solving, we obtain the minimum orce F min mg m 6 9.8 6 Fmin 4.9 0 N. M 0.38 88 s F m M No riction min N EXAMPLE 3.8: Two blocks, stacked one on the other, slide on a rictionless, horizontal surace. The surace between the two blocks is rough, however, with a coeicient o static riction equal to 0.47. (a) I a horizontal orce F is applied to the 5.0-kg bottom m block (m ), what is the maximum value F can have beore the.0-kg top block (m ) begins to slip? F m 6

(b) I the mass o the top block is increased, does the maximum value o F increase, decrease, or stay the same? Explain. Answer: max F m m The maximum rictional orce that the.0-kg block can experience is max, where max m g, that is max = (0.47)(.0 kg)(9.8 ms - ) = 9. N. The acceleration o the -kg block, a = max / m = 4.6 ms -. There will be no slipping when the two block move as one. That is the 5-kg block moves with the same acceleration 4.606 ms -. According to Newton s second law, we have F max = m a, hence F = max + m a = 9. N + (5.0 kg)( 4.6 ms - ) = 3.3 N. We observe that the horizontal orce F m g mg g( m m ). The horizontal orce F increases with the mass o the top block. max EXAMPLE 3.9: A 7.96 kg block rests on a plane inclined at to the horizontal, as shown in igure. The coeicient o static riction is 0.5, while the coeicient o kinetic riction is 0.5. (a) What is the minimum orce F, parallel to the plane, which can prevent the block rom slipping down the plane? (b) What is the minimum orce F that will start the block moving up the plane? (c) What orce F is required to move the block up the plane at constant velocity? Answer: a) When F is at its minimum, s is directed up-hill and assumes its maximum value: F s, max s N smg cos. Thus or the block mgsin Fmin s mg cos 0, which gives F mg sin cos min o s 7.96 9.8sin 0.5 cos.4 N 7

b) To start moving the block up the plane, we must have F mgsin s mg cos 0, which gives the minimum value o F required: F mg sin cos min s 7.96 9.8sin 0.5 cos 47. 3 N c) In this case the block is already in motion, so we should replace s with k in calculation the rictional orce. Thus F mgsin kmg cos ma 0, which gives the value o F: F mg sin cos 7.96 9.8 sin 0.5 cos 40.N k 3.4 Equilibrium Applications o Newton s Law o Motion An object either at rest or moving with a constant velocity is said to be in equilibrium The net orce acting on the object is zero Easier to work with the equation in terms o its components: Fx 0 F y 0 EXAMPLE 3.0: A block o mass m = 5 kg hanging rom three cords, ind the tension in three cords. Answer: Equilibrium condition at knot: F = 0 (two eqs.) x-direction F Bx + F Ax = 0 or F B cos 47 F A cos 8 0 () y-direction F Ay + F By +F Cy = 0 or FB sin 47 FA sin 8 mg 0 () From Eqs.() and (), we obtain F A = 04 N, F B = 35 N 8 o 47 o A knot m C B mg 3.5 Non-Equilibrium Applications o Newton s Law o Motion Similar to equilibrium except F ma 8

Use components F x ma x a x or a y may be zero F y ma y EXAMPLE 3.: Two blocks are connected by a string as shown in the igure. The smooth inclined surace makes an angle o 4 o with the horizontal, and the block on the incline has a mass o m = 6.7 kg. Find the mass o the hanging block m that will cause the system to be in equilibrium. Answer: The ree-body diagrams o the blocks m and m are shown as ollows. m = 6.7 kg 4 o m N T T m g m g The down-plane component o the 6.7-kg mass is given by o o m g sin 4 (6.7 kg)(9.8 ms )sin 4 43.9 N For equilibrium, this orce is balanced i the tension N has the same magnitude. On the other hand, the hanging block is also in equilibrium, the weight o it, m g, must balance the tension orce, hence we can write m g sin4 o m g 43.9. m g 4 o The mass o the hanging block is solved as (43.9 N)/g = 4.48 kg. T m g cos 4 o 9

EXAMPLE 3.: Find the acceleration and the tension o the system. Sliding block M Massless massless pulley rictionless Hanging block m Answer: (a) Free body diagram N Normal orce to prevent it rom alling down T T tension Friction= Mg mg (b) About the pulley I m 0 and the pulley is rotating, then T T. In our problem, the pulley is massless and rictionless i.e. m 0, hence T T T. The string slides on the pulley. T R (c) Cord length is ixed (no extension in the string) a a a T (d) Using Newton s second law T Ma ( ) mg T ma ( ) () + () mg ( M m) a m a M m g Mmg T M m 0

EXAMPLE 3.3: Someone exerts a orce F directly up on the axle o the pulley shown in igure. Consider the pulley and string to be massless and the bearing rictionless. Two F objects, m o mass. kg and m o mass.9 kg, are attached as shown to the opposite ends o the string, which passes over the pulley. The object m is in contact with the loor. (a) What is the largest value the orce F may have so that m will remain at rest on the loor? (b) What is the tension in the string i the upward orce F is 0 N? (c) With the tension determined in part (b), what is the acceleration o m? Answer: m a) When the tension o the string greater than the weight o m, m leaves the loor. Thereore the largest value o orce F or m remain at rest on the loor is F.9 9.8 37.4 N b) Mass o the pulley = 0 kg and net orce acts on the pulley = 0 T 0 T 0 T 55 N c) Net orce acts on m = 55. 9.8 = 43.4 N Acceleration o m = 43.4 /. = 36.0 m s m

WORK, ENERGY AND POWER 3.: WORK DONE BY A CONSTANT FORCE Work is done by a orce when the orce moves a body F s Fig. 3. Fig 3. shows a constant orce F displacing a body through a small displacement, s. DEFINITION: The work done on an object by a constant orce F is W ( F cos ) s - ----3. where F is magnitude o the orce, s is the magnitude o the displacement, and is the angle between the orce and the displacement. Unit: Nm@Joule Work can be either positive or negative, depending on whether a component o the orce points in the same direction as the displacement or in the opposite direction. Work is done when a orce F pushes a car through a displacement,s

EXAMPLE 3. Pulling a suitcase on wheels (a)work can be done by a orce F that point at an angle (b)the orce component that points along the displacement is F cos EXAMPLE 3. A orce o 50N is used to pull a crate at 30 0 rom horizontal on a smooth loor. I the distance is 3.0m, calculate the work done by the orce. Solution: W = ( 50N) ( 50N Cos 30 0 )( 3.0m) = 30 J. EXAMPLE 3.3 Calculate work done rom orce versus displacement graph The work done by a variable orce o magnitude F in moving an object through a displacement o magnitude s is equal to the area under the graph o F cos θ versus s. The angle θ is the angle between the orce and displacement vectors. Area under graph rom 0 to 3.0 m 3 N 0 N3.0 m 0 m = 93 J 3

EXAMPLE 3.4: Two locomotive is used to pull a ship using a cable with T = 5000N. Find the total work done by the locomotives i θ is given as 0 0. Each locomotive does work W = Ts cos θ = (5.00 03 N)(.00 03 m) cos 0.0 = 9.40 0 6 J. The net work is then, WT = W =.88 0 7 J 3.: KINETIC ENERGY AND POTENTIAL ENERGY 3..: KINETIC ENERGY AND WORK-ENERGY THEOREM The kinetic energy KE o an object o mass m and speed v is: KE= ----3. The relationship that relates work to the change in kinetic energy is known as work-energy theorem. The work-energy theorem states that the work W done by the net external orce acting on an object equals the dierence between the object s inal kinetic energy KE and initial kinetic energy KE 0 : W=KE -KE 0= ----3.3 The kinetic energy increases when the net orce does positive work and decreases when the net orce does negative work -Kinetic energy is the energy o motion -Work - positive work (work done on) and - negative work (work done by) -In general, work is a measure o energy transer, and energy is the capacity o doing work. 4

Figure: A constant net external orce F acts over a displacement s and does work on the plane. As a result o a work done, the plane s kinetic energy changes. The net orce is the vector sum o all the external orces acting on the plane, and it is assumed to have the same direction as the displacement s. Newton s nd law; F ma Multiplying both sides by the distance s gives: ( F) s mas The term as on the right side can be related to v0 and v by using equation 0 v v as. By substitute this equation into ( F) s mas, ( F) s mv mv 0 Work done by net ext. orce EXAMPLE 3.5 How much work must be done to stop a 50kg car traveling at 05km/h? Solutions The work done on the car is equal to the change in its kinetic energy, and so m s 5 W KE mv mv 0 50 kg 05km h 5.3 0 J 3.6 km h EXAMPLE 3.7 A 85kg load is lited.0m vertically with an acceleration a = 0.60g by a single cable. Determine i. the tension in the cable ii. the net work done on the load iii. iv. the work done by the cable on the load. the work done by the gravity on the load, and e) the inal speed o the load assuming it started rom rest. Solutions: (a) From the ree-body diagram or the load being lited, write Newton s nd law or 5

the vertical direction, with up being positive. F F mg ma 0.60 mg F T T 3.6mg.6 85 kg 9.80 m s 3.4 0 N (b) The net work done on the load is ound rom the net orce. W F d mg d net net o cos 0 0.60 0.60 85 kg 9.80 m s.0 m 3 9.830 J (c) The work done by the cable on the load is W F d mg d o 4 cos 0.60.6 85 kg 9.80 m s.0 m 7.3 0 J cable T (d) The work done by gravity on the load is W mgd mgd o 4 cos80 85 kg 9.80 m s.0 m 6.4 0 J G (e) Use the work-energy theory to ind the inal speed, with an initial speed o 0. W KE KE mv mv v net net 3 9.830 J W v 0 8.3m s m 85 kg EXAMPLE 3.8 A 58-kg skier is coasting down a 5 slope as shown below. A kinetic rictional orce k =70N opposes her motion. Near the top o the slope, the skier s speed is v0=3.6 m/s. Ignoring air resistance, determine the speed v at a point that is displaced 57m downhill. SOLUTION The external orce points along the x axis and is F mgsin 5 k (58kg)(9.8m / s)(sin 5) 70N 70N 6

The work done by the net orce is W ( F cos ) s [(70N)cos 0](57m) 9700J From work-energy theorem KE W KE0 9700J (58kg)(3.6m / s) 000J So v ( KE ) m (000J ) 58kg 9m / s 3..: POTENTIAL ENERGY WORK DONE BY THE FORCE OF GRAVITY The gravitational potential energy o a body is due to its relative position in a gravitational ield. The drawing depicts a basketball o mass m moving vertically downward, the orce o gravity mg being the only orce acting on the ball. The ball displacement s is downward and has a magnitude o s=h 0 -h To calculate the work, W gravity done on the ball by the orce o gravity, we use W ( F cos ) s With F=mg and θ=0 Wgravity ( mg cos 0)( h h ) mg( h0 h 0 ) -------------3.4 Equation 3.4 valid or any path taken between the initial and inal heights, and not just or the straight down path 7

An object can move along dierent paths in going rom an initial height oh ho to a inal height o h. In each case, the work done by the gravitational orce is the same [ W mg h h )], since the change in vertical distance (h 0 -h ) is the same. gravity ( 0 GRAVITATIONAL POTENTIAL ENERGY To lit a mass m at a constant velocity, the orce required is F=mg When the mass is displaced through a height h, W=Fs=mgh Deinition: The energy that an object o mass m has by virtue o its position relative to the surace o the earth. That position is measured by the height h o the object relative to an arbitrary zero level; PE=mgh ---------3.5 The work done increases the gravitational potential energy o the mass. Hence, when a mass m is raised through a height h, increase in gravitational potential energy is mgh. When a mass m drops through a vertical distance o h, the gravitational potential energy decreases by mgh. 8

From igure above, the pile driver contains a massive hammer that is raised to a height h and then dropped. As a result, the hammer has the potential to do the work o driving the pile into the ground. The greater the height o the hammer, the greater is the potential or doing work, and the greater is the gravitational potential energy. EXAMPLE 3.9 By how much does the gravitational potential energy o a 64 kg high jumper change i his center o mass (body) rises about 4.0m during the jump? Solutions: Subtract the initial gravitational PE rom the inal gravitational PE. PE mgy mgy mg y y 3 64 kg 9.8 m s 4.0 m.5 0 J grav EXAMPLE 3.0 A.60m tall person lits a.0kg book rom the ground so it is.0m above the ground. What is the potential energy o the book relative to a) the ground, and b) the top o the person s head? C) How is the work done by the person related to the answers in parts a) and b)? Solutions: (a) (b) (c) Relative to the ground, the PE is given by PE mg y y.0 kg 9.80 m s.0 m 45.3 J G book ground Relative to the top o the person s head, the PE is given by PE mg y y h.0 kg 9.80m s 0.60 m J G book head The work done by the person in liting the book rom the ground to the inal height is the same as the answer to part (a), 45.3 J. In part (a), the PE is calculated relative to the starting location o the application o the orce on the book. The work done by the person is not related to the answer to part (b). 9

ELASTIC POTENTIAL ENERGY (Giancoli 6 th ed. Pg 47) Elastic potential energy: PE = ½ kx EXAMPLE 3. A 00kg car rolling on a horizontal surace has speed v = 65km/h when it strikes a horizontal coiled spring and is brought to rest in a distance o.m. What is the constant o the spring? Solutions: Assume that all o the kinetic energy o the car becomes PE o the compressed spring. mv 00 kg 65km h 4 mv kx k x. m m s 3.6 km h 8. 0 N m 3.3: THE CONSERVATION OF ENERGY The principle o conservation o energy is energy can either be created or destroyed, but only can be converted rom one orm to another. Whenever energy is transormed rom one orm to another, it is ound that no energy is gained or lost in the process, the total o all the energies beore the process is equal to the total o the energies ater process. 0

Conservative versus nonconservative orces Deinition o conservative orce. A orce is conservative when the work it does on a moving object is independent o the path between the object s initial and inal positions.. A orce is conservative when it does no net work on an object moving around a closed path, starting and inishing at the same point. Example: Gravitational orce, elastic spring orce and electric orce Figure: A roller coaster track is an example o a closed path. Deinition o nonconservative orce. A orce is nonconservative i the work it does on an object moving between points depends on the path o the motion between the points. For a closed path, the total work done by a nonconservative orce is not zero Example: Friction orce, air resistance, tension, normal orce, propulsion orce o a rocket Work done by the net external orce; W W c W nc According to the work-energy theorem, the work done by the external orce is equal to the change in the object s kinetic energy; Wc Wnc mv mv0 I the only conservative orce acting is the gravitational orce, then W c W gravity mg ( h 0 h ) Hence Wnc mv mv0 mgh mgh0 ------3.6 In terms o kinetic and potential energies, we ind Wnc KE KE PE PE KE PE 0 0 -----3.7

3.4: THE CONSERVATION OF MECHANICAL ENERGY (WORK-ENERGY THEOREM) The total mechanical energy E is ormed by combining the concepts o kinetic energy and gravitational energy. The principle o conservation o mechanical energy is the total mechanical energy (E=KE+PE) o an object remains constant as the object moves, provided that the net work done by external nonconservative orces is zero. The work energy theorem can be expressed in terms o the total mechanical energy: W ( KE KE ) ( PE PE ) W nc nc ( KE PE 0 ) ( KE 0 PE @ W nc =E -E 0 ---------- 3.8 Suppose that the net work W done by external nonconservative orces is zero, so W=0. So, E mv E 0 E E 0 mgh mv0 mgh0 --------3.9 0 0 ) Figure above shows the transormations o energy or a bobsled run, assuming that nonconservative orces, such as riction and wind resistance, can be ignored. Kinetic and potential energy can be in converted, while the total mechanical energy remains constant. EXAMPLE 3. A novice skier, starting rom rest, slides down a rictionless 35 0 incline whose vertical height is 85m. How ast is she going when she reaches the bottom? Solutions:

The orces on the skier are gravity and the normal orce. The normal orce is perpendicular to the direction o motion, and so does no work. Thus the skier s mechanical energy is conserved. Subscript represents the skier at the top o the hill, and subscript represents the skier at the bottom o the hill. The ground is the zero location or PE y 0. We have v 0, y 85 m, and y 0 (bottom o the hill). Solve or v, the speed at the bottom. mv mgy mv mgy 0 mgy mv 0 v gy 9.80m s 85 m 60.m s 35mi h 3.5: NONCONSERVATIVE FORCES AND THE WORK ENERGY THEOREM In this case, the inal and initial total energies is equal to W; W E E 0 EXAMPLE 3.3 A 45g baseball is dropped rom a tree 3.0 m above the ground. a) With what speed will it hit the ground i air resistance could be ignored? b) I it actually hits the ground with a speed o 8.0m/s, what is the average orce o air resistance exerted on it? Solutions: (a) Apply energy conservation with no non-conservative work. Subscript represents the ball as it is dropped, and subscript represents the ball as it reaches the ground. The ground is the zero location or gravitational PE. We have v 0, y 3.0 m, and y 0. Solve or v. E E mv mgy mv mgy mgy mv v gy 9.80 m s 3.0 m 6.0 m s (b) Apply energy conservation, but with non-conservative work due to riction included. The work o done by riction will be given by W F d cos80, since the orce o riction is in the NC r opposite direction as the motion. The distance d over which the rictional orce acts will be the 3.0 m distance o all. With the same parameters as above, and v 8.00m s, solve or the orce o riction. W E E F d mv mgy mv mgy F d mgy mv nc r r y v 8.00 m s 3.0 m F m g r d d 0.45 kg 9.80 m s.06 N 3

3.6 POWER Average power, P is deined as the average rate o doing work work W P = ------3.0 time t So, W F. s t t P=Fv (v is speed) -----3. For example, when a machine does more work in a shorter time, the power o the machine is higher. SI unitwatt(w) @ joule per second kw=000w Energy or work done sometimes also measured in kilowatt-hour (kwh) kwh=(x0 3 )(60x60)J=3.6X0 3 J EXAMPLE A 000kg elevator carries a maximum load o 800kg. A Constant rictional orce o 4000N retards its motion upward. What minimum power must the motor deliver to lit the ully loaded elevator at a constant speed o 3.0m/s? Solution: T = Mg Where M is the total mass = 800 kg. Thereore, T = + Mg = 4000N + (800kg)(9.8m/s ). =.6 x 0 4 N. using P = F V, P = TV P =.6 x 0 4 N x 3.0 m/s = 6.48 x 0 4 W 4