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  4. 5.3 kinetic energy and the work

0.2 5.3 kinetic energy and the work-energy theorem  (Page 3/6)

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Determining the work to accelerate a package

Suppose that you push on the 30.0-kg package in [link] with a constant force of 120 N through a distance of 0.800 m, and that the opposing friction force averages 5.00 N.

(a) Calculate the net work done on the package. (b) Solve the same problem as in part (a), this time by finding the work done by each force that contributes to the net force.

Strategy and Concept for (a)

This is a motion in one dimension problem, because the downward force (from the weight of the package) and the normal force have equal magnitude and opposite direction, so that they cancel in calculating the net force, while the applied force, friction, and the displacement are all horizontal. (See [link] .) As expected, the net work is the net force times distance.

Solution for (a)

The net force is the push force minus friction, or F net = 120 N – 5 . 00 N = 115 N size 12{F rSub { size 8{"net"} } " = 120 N – 5" "." "00 N = 115 N"} {} . Thus the net work is

W net = F net d = 115 N 0.800 m = 92.0 N ⋅ m = 92.0 J. alignl { stack { size 12{W rSub { size 8{"net"} } =F rSub { size 8{"net"} } d= left ("115"`N right ) left (0 "." "800"`m right )} {} #" "="92" "." 0`N cdot m="92" "." 0`J "." {} } } {}

Discussion for (a)

This value is the net work done on the package. The person actually does more work than this, because friction opposes the motion. Friction does negative work and removes some of the energy the person expends and converts it to thermal energy. The net work equals the sum of the work done by each individual force.

Strategy and Concept for (b)

The forces acting on the package are gravity, the normal force, the force of friction, and the applied force. The normal force and force of gravity are each perpendicular to the displacement, and therefore do no work.

Solution for (b)

The applied force does work.

W app = F app d cos 0º = F app d = 120 N 0.800 m = 96.0 J alignl { stack { size 12{W rSub { size 8{"app"} } =F rSub { size 8{"app"} } d"cos" left (0° right )=F rSub { size 8{"app"} } d} {} #" "= left ("120 N" right ) left (0 "." "800"" m" right ) {} # " "=" 96" "." "0 J" "." {}} } {}

The friction force and displacement are in opposite directions, so that θ = 180º size 12{θ="180"°} {} , and the work done by friction is

W fr = F fr d cos 180º = − F fr d = − 5.00 N 0.800 m = − 4.00 J. alignl { stack { size 12{W rSub { size 8{"fr"} } =F rSub { size 8{"fr"} } d"cos" left ("180"° right )= - F rSub { size 8{"fr"} } d} {} #" "= - left (5 "." "00 N" right ) left (0 "." "800"" m" right ) {} # ital " "= - 4 "." "00" J "." {}} } {}

So the amounts of work done by gravity, by the normal force, by the applied force, and by friction are, respectively,

W gr = 0, W N = 0, W app = 96.0 J, W fr = − 4.00 J. alignl { stack { size 12{W rSub { size 8{"gr"} } =0,} {} #W rSub { size 8{N} } =0, {} # W rSub { size 8{"app"} } ="96" "." 0" J," {} #W rSub { size 8{"fr"} } = - 4 "." "00"" J" "." {} } } {}

The total work done as the sum of the work done by each force is then seen to be

W total = W gr + W N + W app + W fr = 92 .0 J . size 12{W rSub { size 8{"total"} } =W rSub { size 8{"gr"} } +W rSub { size 8{N} } +W rSub { size 8{"app"} } +W rSub { size 8{"fr"} } ="92" "." 0" J"} {}

Discussion for (b)

The calculated total work W total size 12{W rSub { size 8{"total"} } } {} as the sum of the work by each force agrees, as expected, with the work W net size 12{W rSub { size 8{"net"} } } {} done by the net force. The work done by a collection of forces acting on an object can be calculated by either approach.

Determining speed from work and energy

Find the speed of the package in [link] at the end of the push, using work and energy concepts.

Strategy

Here the work-energy theorem can be used, because we have just calculated the net work, W net size 12{W rSub { size 8{"net"} } } {} , and the initial kinetic energy, 1 2 m v 0 2 size 12{ { {1} over {2} } ital "mv" rSub { size 8{0} rSup { size 8{2} } } } {} . These calculations allow us to find the final kinetic energy, 1 2 mv 2 size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } } {} , and thus the final speed v size 12{v} {} .

Solution

The work-energy theorem in equation form is

W net = 1 2 mv 2 − 1 2 m v 0 2 . size 12{W rSub { size 8{"net"} } = { {1} over {2} } ital "mv" rSup { size 8{2} } - { {1} over {2} } ital "mv" rSub { size 8{0} rSup { size 8{2} } } "." } {}

Solving for 1 2 mv 2 size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } } {} gives

1 2 mv 2 = W net + 1 2 m v 0 2 . size 12{ { {1} over {2} } ital "mv""" lSup { size 8{2} } =w rSub { size 8{ ital "net"} } + { {1} over {2} } ital "mv""" lSub { size 8{0} } "" lSup { size 8{2} } "." } {}

Thus,

1 2 mv 2 = 92 . 0 J + 3 . 75 J = 95. 75 J. size 12{ { {1} over {2} } ital "mv" rSup { size 8{2} } ="92" "." 0`J+3 "." "75"`J="95" "." "75"`J} {}

Solving for the final speed as requested and entering known values gives

v = 2 (95.75 J) m = 191.5 kg ⋅ m 2 /s 2 30.0 kg = 2.53 m/s.

Discussion

Using work and energy, we not only arrive at an answer, we see that the final kinetic energy is the sum of the initial kinetic energy and the net work done on the package. This means that the work indeed adds to the energy of the package.

Work and energy can reveal distance, too

How far does the package in [link] coast after the push, assuming friction remains constant? Use work and energy considerations.

Strategy

We know that once the person stops pushing, friction will bring the package to rest. In terms of energy, friction does negative work until it has removed all of the package’s kinetic energy. The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement; hence, this gives us a way of finding the distance traveled after the person stops pushing.

Solution

The normal force and force of gravity cancel in calculating the net force. The horizontal friction force is then the net force, and it acts opposite to the displacement, so θ = 180º . To reduce the kinetic energy of the package to zero, the work W fr by friction must be minus the kinetic energy that the package started with plus what the package accumulated due to the pushing. Thus W fr = − 95 . 75 J . Furthermore, W fr = f d ′ cos θ = – f d ′ , where d ′ is the distance it takes to stop. Thus,

d ′ = − W fr f = − − 95.75 J 5.00 N , size 12{ { {d}} sup { ' }= - { {W rSub { size 8{"fr"} } } over {f} } = - { { - "95" "." "75"`J} over {5 "." "00 N"} } } {}

and so

d ′ = 19 .2 m . size 12{ { {d}} sup { ' }="19" "." 2" m"} {}

Discussion

This is a reasonable distance for a package to coast on a relatively friction-free conveyor system. Note that the work done by friction is negative (the force is in the opposite direction of motion), so it removes the kinetic energy.

Some of the examples in this section can be solved without considering energy, but at the expense of missing out on gaining insights about what work and energy are doing in this situation. On the whole, solutions involving energy are generally shorter and easier than those using kinematics and dynamics alone.

Section summary

  • The net work W net is the work done by the net force acting on an object.
  • Work done on an object transfers energy to the object.
  • The translational kinetic energy of an object of mass m moving at speed v is KE = 1 2 mv 2 size 12{"KE"= { {1} over {2} } ital "mv" rSup { size 8{2} } } {} .
  • The work-energy theorem states that the net work W net size 12{W rSub { size 8{"net"} } } {} on a system changes its kinetic energy, W net = 1 2 mv 2 − 1 2 m v 0 2 .

Conceptual questions

The person in [link] does work on the lawn mower. Under what conditions would the mower gain energy? Under what conditions would it lose energy?

A person pushing a lawn mower with a force F. Force is represented by a vector making an angle theta below the horizontal and distance moved by the mover is represented by vector d. The component of vector F along vector d is F cosine theta. Work done by the person, W, is equal to F d cosine theta.

Work done on a system puts energy into it. Work done by a system removes energy from it. Give an example for each statement.

Problems&Exercises

How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0-kg sprinter running at 10.0 m/s?

(a) Calculate the force needed to bring a 950-kg car to rest from a speed of 90.0 km/h in a distance of 120 m (a fairly typical distance for a non-panic stop). (b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a).

A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object without damage to the body of the car. The bumper cushions the shock by absorbing the force over a distance. Calculate the magnitude of the average force on a bumper that collapses 0.200 m while bringing a 900-kg car to rest from an initial speed of 1.1 m/s.

2 . 8 × 10 3 N size 12{2 "." 8 times "10" rSup { size 8{3} } " N"} {}

Using energy considerations, calculate the average force a 60.0-kg sprinter exerts backward on the track to accelerate from 2.00 to 8.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.

102 N
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Source:  OpenStax, Unit 5 - work and energy. OpenStax CNX. Jan 02, 2016 Download for free at https://legacy.cnx.org/content/col11946/1.1
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