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1.2 Kinetic energy and the work-energy theorem  (Page 5/10)

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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.
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Source:  OpenStax, Work and energy. OpenStax CNX. Nov 09, 2015 Download for free at http://legacy.cnx.org/content/col11902/1.1
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