A Mother Pushes a Baby Stroller 10 Meters by Applying 40 Newtons of Force. How Much Work Was Done?

5.ii Work (ESCMB)

We cover different topics in dissimilar chapters in unlike grades but that doesn't mean that they are non related. In fact, it is very important to note that all of the different topics related to mechanics (forces, mechanical free energy, momentum, rectilinear movement) actually form a consistent moving-picture show of the same physical organisation. There take been examples where we've shown the same results using two methods, for instance determining speed or velocity using equations of motion or conservation of mechanical energy. Learning about work will aid us necktie everything we've learnt most previously together.

Work will let united states of america to connect energy transfer to forces, which nosotros accept already linked to momentum and the equations of move. When a force tends to act in or against the management of move of an object we say that the strength is doing work on the object. Specifically, work is defined mathematically in terms of the force and the deportation of the object.

Piece of work

When a strength acts in or against the management of motility of an object, work is done on the object.

\[W=F\Delta ten\cos\theta\]

This means that in gild for piece of work to exist washed, an object must have its position changed past an amount \(\Delta \vec{x}\) while a forcefulness, \(\vec{F}\), is acting on it such that at that place is some non-zero component of the force in the direction of the displacement. Work is calculated as:

\[W=F\Delta ten\cos\theta\]

Where:

• \(W\) = work in joules (J)
• \(F\) = magnitude of force for which we are calculating work in newtons (Northward)
• \(\Delta x\) = magnitude of deportation in metres (1000)
• \(\theta\) = bending betwixt the force and the management of displacement

Note that:

• \(F\cos\theta\) is the magnitude of the component of \(\vec{F}\) in the direcion of \(\Delta \vec{x}\). If \(\theta > xc^{o}\) then the component is parallel to the direction of displacement but point in the opposite management and the force is opposing the move.
• Nosotros will deal with constant forces not forces that are continuously changing in this affiliate. All of the principles we have covered are applicable when forces are non constant.

Figure v.1: The strength \(\vec{F}\) causes the object to be displaced past \(\Delta \vec{x}\) at bending θ.

Important: \(\cos\theta\) tells you the relative direciton of the force and the displacment which is important. If the component of the strength along the direction of the displacement is contrary in direction to the displacement then the sign of the displacement vector and force vector volition be dissimilar. This is regardless of which direction was called as a positive direction.

Let us wait at some examples to understand this properly. In the images beneath the grey dot represents an object. A forcefulness, \(\vec{F}\), acts on the object. The object moves through a displacement, \(\Delta \vec{x}\). What is the sign of the work washed in each case?

At \(\theta=xc^\circ\) the magnitude of the component of the force in the direction of the displacement is zero then no piece of work is washed as \(\cos(xc)=0\).

At \(\theta>90^\circ\) the magnitude of the component of the strength along the line of the displacement is opposite in direction to the displacement so negative work is washed, \(\cos\theta < 0\).

It is very important to annotation that for work to be done there must be a component of the practical force in the management of motion. Forces perpendicular to the direction of motility exercise no work.

Figure five.2: The force \(\vec{F}\) causes the object to be displaced by \(\Delta \vec{10}\) at angle θ.

It is simply the direction of the strength on the object that matters and not the direction from the source of the force to the object. In Effigy 5.3 both powerlifters are exerting an upwards strength on the weights. On the left the weight is being pulled upwards and on the right it is being pushed upwards.

Effigy 5.3: Left: A powerlifter deadlifts a weight. Right: A powerlifter benchpresses a weight.

Weight lifting is a good context in which to think nearly work because it helps to place misconceptions introduced by everyday use of the term `piece of work'. In the two cases in Figure 5.3 anybody would describe moving the weights upwardly as very hard piece of work. From a physics perspective, equally the powerlifters elevator the weight they are exerting a strength in the direction of the displacement so positive piece of work is done on the weights.

Consider the strongman walking in Effigy 5.4. He carries two very heavy sleds as far equally he can in a competition. What work is the human doing on the sleds and why?

Figure 5.iv: A strongman carries heavy sleds as far as possible in a competition.

Almost people would say he is working very hard because the sleds are heavy to carry but from a physics perspective he is doing no piece of work on the sleds. The reason that he does no piece of work is because the force he exerts is direct upward to rest the forcefulness of gravity and the displacement is in the horizontal direction. Therefore in that location is no component of the force in the direction of deportation (\(\theta=90^\circ\)) and no piece of work done.

His muscles do need to employ their energy reserves to maintain the force to balance gravity. That does not result in energy transfer to the sleds.

Is piece of work washed?

Decide whether or not work is done in the post-obit situations. Call back that for piece of work to be done a force must be applied in the direction of motion and there must be a displacement. Identify which two objects are interacting, what the activeness-reaction pairs of forces are and why the force described is or isn't doing work.

1. Max pushes confronting a wall and becomes tired.

2. A book falls off a table and free falls to the ground.

3. A rocket accelerates through space.

4. A waiter holds a tray total of plates above his head with 1 arm and carries information technology direct across the room at constant speed.

1. Max pushes against a wall and becomes tired. Answer: No work is done considering there is no displacement.

2. A book falls off a table and gratuitous falls to the ground. Answer: Yes piece of work is done because there is a displacement in the direction of the strength of gravity.

3. A rocket accelerates through infinite. Answer: Yep piece of work is done because in that location is a internet forcefulness interim for there to be a net acceleration. If there is an dispatch and then at that place is a deportation.

4. A waiter holds a tray full of meals to a higher place his head with one arm and carries it directly across the room at constant speed. (Conscientious! This is a catchy question.) Reply: No work is done considering at that place is no net forcefulness in the direction of the displacement.

For each of the above pictures, the force vector is acting in the aforementioned direction every bit the displacement vector. As a upshot, the angle \(\theta =0°\) because at that place is no deviation in bending between the management of applied forcefulness and the management of deportation.

The piece of work done by a forcefulness can then be positive or negative. This sign tells united states nearly the direction of the energy transfer. Piece of work is a scalar so the sign should not exist misinterpreted to hateful that work is a vector. Piece of work is divers equally energy transfer, energy is a scalar quantity and the sign indicates whether energy was increased or decreased.

• If\(\vec{F}_{\text{applied}}\) acts or has a component interim in the same direction equally the motion, and so positive work is beingness done. In this instance the object on which the force is applied gains energy.

• If the direction of motion and \(\vec{F}_{\text{applied}}\) are opposite, so negative work is being done. This ways that energy is lost and the object exerting the force gains energy. For example, if yous try to button a car uphill by applying a force up the slope and instead the car rolls downwards the colina y'all are doing negative work on the motorcar. Alternatively, the car is doing positive work on yous!

The everyday use of the word "work" differs from the physics utilize. In physics, just the component of the applied forcefulness that is parallel to the motion does work on an object. So, for example, a person holding upward a heavy book does no piece of work on the book.

As with all concrete quantities, piece of work must have units. Following from the definition, piece of work is measured in \(\text{N·m}\). The name given to this combination of S.I. units is the joule (symbol J).

Worked case 1: Calculating piece of work on a car when speeding up.

A car is travelling along a straight horizontal road. A force of \(\text{500}\) \(\text{N}\) is applied to the car in the direction that information technology is travelling, speeding it up. While it is speeding upward is covers a altitude of \(\text{20}\) \(\text{grand}\). Calculate the work done on the car.

Analyse the question to determine what information is provided

• The magnitude of the force applied is \(F=\text{500}\text{ Due north}\).

• The altitude moved is \(\Delta x\) = \(\text{20}\) \(\text{thousand}\).

• The applied strength and distance moved are in the same direction. Therefore, the angle between the force and displacement is \(\theta=0^\circ\).

These quantities are all in SI units, so no unit conversions are required.

Analyse the question to determine what is being asked

• We are asked to observe the piece of work washed on the motorcar. We know from the definition that piece of work done is \(Due west={F}\Delta x\cos\theta\).

Adjacent nosotros substitute the values and calculate the work washed

\begin{align*} W& = {F}\Delta x\cos\theta\\ & = \left(500\correct)\left(twenty\right) \left(\cos 0\right)\\ & = \left(500\right)\left(xx\right) \left(1\correct)\\ & = \text{10 000}\text{ J} \finish{align*}

Think that the respond must be positive, as the practical strength and the deportation are in the same management. In this case, the machine gains kinetic free energy.

Worked example 2: Calculating work on the auto while braking

The aforementioned auto at present slows downwardly when a force of \(\text{300}\) \(\text{N}\) is practical opposite to the direction of motion while it travels \(\text{25}\) \(\text{k}\) forward. Calculate the work done on the car.

Analyse the question to determine what information is provided

• The magnitude of the forcefulness practical is \(F=\text{300}\text{ Northward}\).

• The altitude moved is \(\Delta x\) = \(\text{25}\) \(\text{m}\).

• The applied force and distance moved are in the reverse management. Therefore, the bending between the force and displacement is \(\theta=180^\circ\).

These quantities are all in the correct units, so no unit of measurement conversions are required.

Analyse the question to determine what is being asked

• We are asked to find the work washed on the auto. We know from the definition that work done is \(W={F}\Delta x\cos\theta\)

Next we substitute the values and calculate the piece of work done

\brainstorm{marshal*} W& = {F}\Delta x\cos\theta\\ & = \left(300\right)\left(25\correct) \left(\cos 180\right)\\ & = \left(300\correct)\left(25\right) \left(-ane\right)\\ & = -\text{7 500}\text{ J} \end{marshal*}

Note that the respond must be negative as the applied force and the displacement are in reverse directions. This ways that the free energy is beingness lost past the motorcar. This may exist energy lost every bit heat to the environment. Energy conservation still holds, the energy has just been transferred to a larger arrangement that contains the car.

What happens when the applied strength and the motion are not parallel? By using the formula \(W=F\Delta 10\cos\theta\) , we are actually calculating the component of the practical force in the direction of motion. Note that the component of the force perpendicular to the direction of movement does no work.

Worked instance 3: Calculating work done on a box pulled at an bending.

Calculate the work done on a box, if it is pulled \(\text{v}\) \(\text{m}\) forth the footing past applying a force of \(F=\text{20}\text{ N}\) at an angle of \(\text{60}\)\(\text{°}\) to the horizontal.

Analyse the question to determine what information is provided

• The force practical is \(F\)=\(\text{20}\) \(\text{N}\)

• The altitude moved is \(\Delta x\) = \(\text{5}\) \(\text{m}\) along the ground

• The angle between the applied force and the motion is \(\theta\)=\(\text{60}\)\(\text{°}\)

These quantities are in the correct units so we practice not need to perform any unit conversions.

Analyse the question to determine what is being asked

• Nosotros are asked to find the work done on the box.

Substitute and calculate the work done

Now nosotros tin can summate the work done on the box:

\begin{align*} W& = {F}\Delta x\cos\theta\\ & = \left(xx \right)\left(five \right) \left(\cos sixty\right)\\ & = \text{50}\text{ J} \end{marshal*}

Note that the answer is positive as the component of the forcefulness parallel to the direction of movement is in the same direction as the motion.

The work done on the box is \(\text{fifty}\) \(\text{J}\).

Work

Textbook Exercise five.1

A \(\text{x}\) \(\text{Due north}\) strength is applied to push a cake across a frictionless surface for a displacement of \(\text{5,0}\) \(\text{m}\) to the right. The block has a weight \(\vec{F}_{g }\) of \(\text{20}\) \(\text{N}\). Determine the piece of work done by the following forces: normal forcefulness, weight \(\vec{F}_{g }\), applied strength.

Normal forcefulness acts perpendicular and therefore does no piece of work.

The weight acts perpendicular to the direction of motion and does no piece of work.

The applied forcefulness: \begin{align*} W_{\text{applied}}&= F\Delta ten \cos\theta \\ &= (ten)(5) \\ & = 50~\text{J} \finish{marshal*}

Normal strength:0 J, weight: 0 J, applied force: fifty J.

A \(\text{10}\) \(\text{North}\) frictional forcefulness slows a moving block to a end later a deportation of \(\text{5,0}\) \(\text{m}\) to the right. The block has a weight of \(\text{20}\) \(\text{N}\) Determine the piece of work done by the following forces: normal force, weight, frictional force.

Normal force acts perpendicular and therefore does no work.

The weight acts perpendicular to the direction of motion and does no work.

The frictional strength: \brainstorm{align*} W_{friciton}&= F\Delta x \cos\theta \\ &= (x)(5)\cos(180) \\ & = -50~\text{J} \end{align*}

Normal force:0 J, weight: 0 J, frictional strength: -fifty J.

A \(\text{ten}\) \(\text{N}\) strength is applied to button a block across a frictional surface at constant speed for a displacement of \(\text{5,0}\) \(\text{m}\) to the right. The block has a weight of \(\text{xx}\) \(\text{N}\) and the frictional force is \(\text{10}\) \(\text{N}\). Determine the work done past the following forces: normal force, weight, applied force and frictional strength.

Normal force acts perpendicular and therefore does no work.

The weight acts perpendicular to the direction of motility and does no work.

The frictional force: \begin{align*} W_{friciton}&= F\Delta x \cos\theta \\ &= (10)(5)\cos(180) \\ & = -l~\text{J} \finish{align*}

The practical forcefulness: \begin{align*} W_{\text{applied}}&= F\Delta x \cos\theta \\ &= (ten)(v) \\ & = 50~\text{J} \end{align*}

Normal strength:0 J, weight: 0 J, frictional force: -50 J, practical force: 50 J.

An object with a weight of \(\text{xx}\) \(\text{Northward}\) is sliding at constant speed across a frictionless surface for a deportation of \(\text{five}\) \(\text{m}\) to the correct. Determine if there is any work done.

No work is done as all the forces act perpendicular to the management of displacement.

An object with a weight of \(\text{20}\) \(\text{Northward}\) is pulled upward at abiding speed past a \(\text{20}\) \(\text{North}\) force for a vertical displacement of \(\text{5}\) \(\text{m}\). Determine if there is any work done.

The practical force: \begin{align*} W_{\text{applied}}&= F\Delta 10 \cos\theta \\ &= (20)(five) \\ & = 100~\text{J} \finish{marshal*}

The gravitational force: \begin{align*} W_{gravitational}&= F\Delta ten \cos\theta \\ &= (20)(v)\cos(180) \\ &= -(xx)(5) \\ & = -100~\text{J} \stop{align*}

Work done by applied forcefulness is 100 J, work done by gravity is -100 J

Before beginning its descent, a roller coaster is always pulled up the get-go hill to a high initial height. Work is done on the roller coaster to achieve this initial superlative. A coaster designer is considering three different incline angles of the loma at which to drag the \(\text{2 000}\) \(\text{kg}\) auto railroad train to the summit of the \(\text{60}\) \(\text{chiliad}\) high hill. In each case, the force practical to the car will be practical parallel to the hill. Her critical question is: which angle would require the least work? Analyse the data, determine the work done in each case, and answer this critical question.

Angle of Incline	Applied Force	Distance	Work
\(\text{35}\)\(\text{°}\)	\(\text{1,1} \times \text{ten}^{\text{4}}\) \(\text{N}\)	\(\text{100}\) \(\text{m}\)
\(\text{45}\)\(\text{°}\)	\(\text{1,3} \times \text{10}^{\text{4}}\) \(\text{Northward}\)	\(\text{ninety}\) \(\text{chiliad}\)
\(\text{55}\)\(\text{°}\)	\(\text{1,5} \times \text{10}^{\text{4}}\) \(\text{N}\)	\(\text{fourscore}\) \(\text{m}\)

Angle of Incline	Applied Force	Distance	Piece of work
\(\text{35}\)\(\text{°}\)	\(\text{1,1} \times \text{x}^{\text{iv}}\) \(\text{Due north}\)	\(\text{100}\) \(\text{chiliad}\)	\(\text{9,01} \times \text{10}^{\text{5}}\) \(\text{J}\)
\(\text{45}\)\(\text{°}\)	\(\text{ane,3} \times \text{10}^{\text{iv}}\) \(\text{N}\)	\(\text{90}\) \(\text{m}\)	\(\text{8,27} \times \text{10}^{\text{v}}\) \(\text{J}\)
\(\text{55}\)\(\text{°}\)	\(\text{1,5} \times \text{ten}^{\text{4}}\) \(\text{North}\)	\(\text{80}\) \(\text{m}\)	\(\text{vi,88} \times \text{10}^{\text{5}}\) \(\text{J}\)

An bending of 55° requires the least amount of work to exist done.

A traveller carries a \(\text{150}\) \(\text{N}\) suitcase up four flights of stairs (a total height of \(\text{12}\) \(\text{m}\)) and then pushes it with a horizontal force of \(\text{threescore}\) \(\text{North}\) at a abiding speed of \(\text{0,25}\) \(\text{m·s$^{-i}$}\) for a horizontal distance of \(\text{fifty}\) \(\text{one thousand}\) on a frictionless surface. How much work does the traveller do on the suitcase during this entire trip?

1800 J up stairs and 3000 J along passage.

A parent pushes downwardly on a pram with a force of \(\text{50}\) \(\text{N}\) at an angle of \(\text{xxx}\)\(\text{°}\) to the horizontal. The pram is moving on a frictionless surface. If the parent pushes the pram for a horizontal distance of \(\text{30}\) \(\text{chiliad}\), how work is done on the pram?

\brainstorm{align*} Westward&= F\Delta x \cos\theta \\ &= (fifty)(thirty)\cos(30)\\ & = \text{1 299,00}\text{ J} \end{marshal*} 1,30 kJ

How much work is washed by the strength required to raise a \(\text{two 000}\) \(\text{Northward}\) lift 5 floors vertically at a constant speed? The vertical distance between floors is \(\text{5}\) \(\text{m}\) loftier.

\begin{align*} W&= F\Delta x \cos\theta \\ &= (2000)(5\times v) \cos(0) \\ &= 2000 \times 25 \\ &= \text{fifty 000}\text{ J} \terminate{align*} 50 000 J

A student with a mass of \(\text{60}\) \(\text{kg}\) runs up three flights of stairs in \(\text{15}\) \(\text{southward}\), roofing a vertical distance of \(\text{10}\) \(\text{m}\). Determine the amount of work done by the student to elevate her torso to this height.

\brainstorm{align*} Westward&= F\Delta 10 \cos\theta \\ &= mg \Delta 10 \cos \theta \\ &= (60)(9,8) (x) \\ &= \text{5 880}\text{ J} \end{align*} 5 880 J

Net piece of work (ESCMC)

We have merely looked at a single force interim on an object. Sometimes more than one force acts at the same time (we dealt with this in Grade xi). We phone call the piece of work washed after taking all the forces into business relationship the internet work washed. If there is only one force acting so the work it does, if any, is the net work done. In this example in that location are ii equivalent approaches we can adopt to finding the net work washed on the object. We can:

• Arroyo i: calculate the piece of work done by each force individually and then sum them taking the signs into account. If one force does positive work and another does the same amount of piece of work but information technology is negative then they cancel out.
• Approach 2: calculate the resultant forcefulness from all the forces interim and calculate the work done using the resultant force. This will be equivalent to Approach i. If the resultant force parallel to the direction of move is nix, no cyberspace work will be done.

Call up that work done tells you lot about the energy transfer to or from an object past means of a strength. That is why we tin can accept zero cyberspace piece of work washed fifty-fifty if multiple large forces are acting on an object. Forces that issue in positive work increment the free energy of the object, forces that result in negative work reduce the energy of an object. If as much free energy is transferred to an object every bit is transferred away and so the final result is that the object gains no energy overall.

Worked example four: Approach 1, computing the internet work on a car

The same car is at present accelerating forward, but friction is working against the motion of the automobile. A forcefulness of \(\text{300}\) \(\text{N}\) is practical forward on the automobile while it is travelling \(\text{20}\) \(\text{yard}\) frontward. A frictional force of \(\text{100}\) \(\text{N}\) acts to oppose the movement. Calculate the net work done on the car.

Only forces with a component in the plane of motion are shown on the diagram. No work is done by \({F}_{g}\) or \({F}_{\text{normal}}\) as they deed perpendicular to the management of movement.

Analyse the question to determine what data is provided

• The force practical is \(F_{\text{applied}}\)=\(\text{300}\) \(\text{N}\) forwards.

• The strength of friction is \(F_{\text{friction}}\)=\(\text{100}\) \(\text{N}\) opposite to the direction of motion.

• The distance moved is \(\Delta x\) = \(\text{20}\) \(\text{thousand}\).

• The applied force and distance moved are in the aforementioned plane so nosotros can calculate the piece of work done by the applied forwards force and the work done by the forcefulness of friction backwards.

These quantities are all in the correct units, so no unit of measurement conversions are required.

Analyse the question to determine what is being asked

• We are asked to observe the net work done on the car. Nosotros know from the definition that piece of work done is \(Due west={F}\)\(\Delta ten\) \(\cos\theta\)

Adjacent we summate the work done by each force.

\begin{align*} W_{\text{applied}}& = {F}_{\text{applied}}\Delta x\cos\theta\\ & = \left(300\right)\left(20\right) \left(\cos 0\correct)\\ & = \left(300\right)\left(20\right) \left(1\correct)\\ & = \text{6 000}\text{ J} \end{align*}\brainstorm{align*} W_{\text{friction}}& = {F}_{\text{friction}}\Delta ten\cos\theta\\ & = \left(100\right)\left(20\right) \left(\cos 180\right)\\ & = \left(100\right)\left(xx\right) \left(-1\correct)\\ & = -\text{2 000}\text{ J} \end{align*}\begin{align*} {Due west}_{\text{net}}& = {West}_{\text{applied}} + {W}_{\text{friction}} \\ & = \left(6000\correct) + \left(-2000\right) \\ & = \text{4 000}\text{ J} \end{align*}

The answer shown in this worked example shows that although free energy has been lost by the car to friction, the total work done on the motorcar has resulted in a net free energy gain. This can be seen by the positive answer.

Equally mentioned before, there is an alternative method to solving the same problem, which is to make up one's mind the net force acting on the auto and to apply this to calculate the work. This means that the vector forces acting in the aeroplane of motion must exist added to go the net force \(\vec{F}_{\text{net}}\). The net force is then practical over the displacement to get the net work \({West}_{\text{net}}\).

Worked case 5: Approach ii, calculating the internet force

The aforementioned car is now accelerating frontwards, but friction is working against the move of the automobile. A force of \(\text{300}\) \(\text{N}\) is practical frontward on the automobile while information technology is travelling \(\text{20}\) \(\text{m}\) forward. A frictional force of \(\text{100}\) \(\text{Due north}\) acts to oppose the motion. Summate the cyberspace work done on the car.

But forces with a component in the plane of motion are shown on the diagram. No work is done past \({F}_{k}\) or \({F}_{\text{normal}}\) as they deed perpendicular to the direction of motion. The net force acting in the plane of move volition be calculated using the non-perpendicular forces.

Analyse the question to determine what information is provided

• The forcefulness applied is \(\vec{F}_{\text{applied}}\)=\(\text{300}\) \(\text{North}\) forrad.

• The forcefulness of friction is \(\vec{F}_{\text{friction}}\)=\(\text{100}\) \(\text{N}\) backwards.

• The distance moved is \(\Delta x\) = \(\text{20}\) \(\text{grand}\).

• The applied forces \(\vec{F}_{\text{applied}}= \text{300}\text{ N}\) and the forcefulness of friction \(\vec{F}_{\text{friction}}= \text{100}\text{ N}\) are in the aforementioned plane every bit the distance moved. Therefore, we tin can add the vectors. As vectors require management, we will say that forward is positive and therefore backward is negative. Annotation, the force of friction is acting at \({180}^{0}\) i.due east. backwards and then is acting in the contrary vector direction i.e. negative.

These quantities are all in the correct units, and then no unit conversions are required.

Analyse the question to determine what is being asked

• We are asked to notice the net work washed on the machine. We know from the definition that work washed is \(W_{\text{cyberspace}}={F}_{\text{net}}\)\(\Delta x \cos \theta\)

We calculate the cyberspace forcefulness acting on the auto, and we convert this into net piece of work.

Outset we draw the strength diagram:

Let forwards (to the left in the picture) be positive. We know that the motility of the automobile is in the horizontal direction so nosotros tin can neglect the force due to gravity, \(\vec{F}_g\), and the normal force, \(\vec{N}\). Note: if the car were on a slope we would need to summate the component of gravity parallel to the slope.

\brainstorm{marshal*} \vec{F}_{\text{net}}& = \vec{F}_{\text{applied}} + \vec{F}_{\text{friction}}\\ & = \left(+300\right) + \left(-100\right)\\ \vec{F}_{\text{net}}& = \text{200}\text{ N}~\text{forwards} \end{align*}

\(\vec{F}_{\text{net}}\) is pointing in the same direction as the displacement, therefore the angle between the force and displacement is \(\theta=0^\circ\).

\begin{align*} {W}_{\text{internet}}& = {F}_{\text{internet}}\Delta 10 \cos \theta \\ & = \left(200\right)\left(20\right)\cos\left(0\right)\\ & = \text{four 000}\text{ J} \stop{align*}

The two different approaches give the same effect but it is very important to care for the signs correctly. The forces are vectors but piece of work is a scalar and then they shouldn't exist interpreted in the aforementioned manner.

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