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W net = W nc + W c , size 12{W rSub { size 8{"net"} } =W rSub { size 8{"nc"} } +W rSub { size 8{c} } } {}

so that

W nc + W c = Δ KE , size 12{W rSub { size 8{"nc"} } +W rSub { size 8{c} } =Δ"KE"} {}

where W nc size 12{W rSub { size 8{"nc"} } } {} is the total work done by all nonconservative forces and W c size 12{W rSub { size 8{c} } } {} is the total work done by all conservative forces.

A person pushing a heavy box up an incline. A force F p applied by the person is shown by a vector pointing up the incline. And frictional force f is shown by a vector pointing down the incline, acting on the box.
A person pushes a crate up a ramp, doing work on the crate. Friction and gravitational force (not shown) also do work on the crate; both forces oppose the person’s push. As the crate is pushed up the ramp, it gains mechanical energy, implying that the work done by the person is greater than the work done by friction.

Consider [link] , in which a person pushes a crate up a ramp and is opposed by friction. As in the previous section, we note that work done by a conservative force comes from a loss of gravitational potential energy, so that W c = Δ PE size 12{W rSub { size 8{c} } = - Δ"PE"} {} . Substituting this equation into the previous one and solving for W nc size 12{W rSub { size 8{"nc"} } } {} gives

W nc = Δ KE + Δ PE. size 12{W rSub { size 8{"nc"} } =Δ"KE"+Δ"PE"} {}

This equation means that the total mechanical energy ( KE + PE ) size 12{ \( "KE + PE" \) } {} changes by exactly the amount of work done by nonconservative forces. In [link] , this is the work done by the person minus the work done by friction. So even if energy is not conserved for the system of interest (such as the crate), we know that an equal amount of work was done to cause the change in total mechanical energy.

We rearrange W nc = Δ KE + Δ PE size 12{W rSub { size 8{"nc"} } =D"KE"+D"PE"} {} to obtain

KE i + PE i + W nc = KE f + PE f . size 12{"KE""" lSub { size 8{i} } +"PE" rSub { size 8{i} } +W rSub { size 8{"nc"} } ="KE""" lSub { size 8{f} } +"PE" rSub { size 8{f} } } {}

This means that the amount of work done by nonconservative forces adds to the mechanical energy of a system. If W nc size 12{W rSub { size 8{"nc"} } } {} is positive, then mechanical energy is increased, such as when the person pushes the crate up the ramp in [link] . If W nc size 12{W rSub { size 8{"nc"} } } {} is negative, then mechanical energy is decreased, such as when the rock hits the ground in [link] (b). If W nc size 12{W rSub { size 8{"nc"} } } {} is zero, then mechanical energy is conserved, and nonconservative forces are balanced. For example, when you push a lawn mower at constant speed on level ground, your work done is removed by the work of friction, and the mower has a constant energy.

Applying energy conservation with nonconservative forces

When no change in potential energy occurs, applying KE i + PE i + W nc = KE f + PE f size 12{"KE""" lSub { size 8{i} } +"PE" rSub { size 8{i} } +W rSub { size 8{"nc"} } ="KE""" lSub { size 8{f} } +"PE" rSub { size 8{f} } } {} amounts to applying the work-energy theorem by setting the change in kinetic energy to be equal to the net work done on the system, which in the most general case includes both conservative and nonconservative forces. But when seeking instead to find a change in total mechanical energy in situations that involve changes in both potential and kinetic energy, the previous equation KE i + PE i + W nc = KE f + PE f size 12{"KE""" lSub { size 8{i} } +"PE" rSub { size 8{i} } +W rSub { size 8{"nc"} } ="KE""" lSub { size 8{f} } +"PE" rSub { size 8{f} } } {} says that you can start by finding the change in mechanical energy that would have resulted from just the conservative forces, including the potential energy changes, and add to it the work done, with the proper sign, by any nonconservative forces involved.

Calculating distance traveled: how far a baseball player slides

Consider the situation shown in [link] , where a baseball player slides to a stop on level ground. Using energy considerations, calculate the distance the 65.0-kg baseball player slides, given that his initial speed is 6.00 m/s and the force of friction against him is a constant 450 N.

A baseball player slides to stop in a distance d. the displacement d is shown by a vector towards the left and frictional force f on the player is shown by a small vector pointing towards the right equal to four hundred and fifty newtons. K E is equal to half m v squared, which is equal to f times d.
The baseball player slides to a stop in a distance d size 12{d} {} . In the process, friction removes the player’s kinetic energy by doing an amount of work fd size 12{ ital "fd"} {} equal to the initial kinetic energy.

Strategy

Friction stops the player by converting his kinetic energy into other forms, including thermal energy. In terms of the work-energy theorem, the work done by friction, which is negative, is added to the initial kinetic energy to reduce it to zero. The work done by friction is negative, because f size 12{f} {} is in the opposite direction of the motion (that is, θ = 180º size 12{q="180"°} {} , and so cos θ = 1 size 12{"cos"θ= - 1} {} ). Thus W nc = fd size 12{W rSub { size 8{"nc"} } = - ital "fd"} {} . The equation simplifies to

1 2 mv i 2 fd = 0 size 12{ { {1} over {2} }  ital "mv" rSub { size 8{i} rSup { size 8{2} } } - ital "fd"=0} {}

or

fd = 1 2 mv i 2 . size 12{ ital "fd"= { {1} over {2} }  ital "mv" rSub { size 8{i} rSup { size 8{2} } } "." } {}

This equation can now be solved for the distance d size 12{d} {} .

Solution

Solving the previous equation for d size 12{d} {} and substituting known values yields

d = mv i 2 2 f = ( 65.0 kg ) ( 6 . 00 m/s ) 2 ( 2 ) ( 450 N ) = 2.60 m. alignl { stack { size 12{d= { { ital "mv" rSub { size 8{i} rSup { size 8{2} } } } over {2f} } } {} #= { { \( "65" "." 0" kg" \) \( 6 "." "00"" m/s" \) rSup { size 8{2} } } over { \( 2 \) \( "450"" N" \) } } {} # " "=" 2" "." "60 m" "." {}} } {}

Discussion

The most important point of this example is that the amount of nonconservative work equals the change in mechanical energy. For example, you must work harder to stop a truck, with its large mechanical energy, than to stop a mosquito.

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Questions & Answers

find the density of a fluid in which a hydrometer having a density of 0.750g/mL floats with 92.0% of its volume submerged.
Neshrin Reply
Uniform speed
Sunday
(a)calculate the buoyant force on a 2.00-L Helium balloon.(b) given the mass of the rubber in the balloon is 1.50g. what is the vertical force on the balloon if it is let go? you can neglect the volume of the rubber.
Neshrin Reply
To Long
Usman
pleaseee. can you get the answer? I can wait till 12
Neshrin
a thick glass cup cracks when hot liquid is poured into it suddenly
Aiyelabegan Reply
because of the sudden contraction that takes place.
Eklu
railway crack has gap between the end of each length because?
Aiyelabegan Reply
For expansion
Eklu
yes
Aiyelabegan
Please i really find it dificult solving equations on physic, can anyone help me out?
Big Reply
sure
Carlee
what is the equation?
Carlee
Sure
Precious
fersnels biprism spectrometer how to determined
Bala Reply
how to study the hall effect to calculate the hall effect coefficient of the given semiconductor have to calculate the carrier density by carrier mobility.
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Nana Reply
find the dimensional equation of work,power,and moment of a force show work?
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Peter
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Okeh
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Nana
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Nana
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Joseph
Michelson Morley experiment
Riya Reply
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Naveed
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Celine
you
Celine
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Bala
Hi
Ahmed
Calculate the final velocity attained, when a ball is given a velocity of 2.5m/s, acceleration of 0.67m/s² and reaches its point in 10s. Good luck!!!
Eklu Reply
2.68m/s
Doc
vf=vi+at vf=2.5+ 0.67*10 vf= 2.5 + 6.7 vf = 9.2
babar
s = vi t +1/2at sq s=58.5 s=v av X t vf= 9.2
babar
how 2.68
babar
v=u+at where v=final velocity u=initial velocity a=acceleration t=time
Eklu
the answer is 9.2m/s
OBERT
express your height in Cm
Emmanuel Reply
my project is Sol gel process how to prepare this process pls tell me
Bala
the dimension of work and energy is ML2T2 find the unit of work and energy hence drive for work?
Emmanuel Reply
KgM2S2
Acquah
Two bodies P and Quarter each of mass 1000g. Moved in the same direction with speed of 10m/s and 20m/s respectively. Calculate the impulse of P and Q obeying newton's 3rd law of motion
Shimolla Reply
kk
Doc
the answer is 0.03n according to the 3rd law of motion if the are in same direction meaning they interact each other.
OBERT
definition for wave?
Doc Reply
A disturbance that travel from one medium to another and without causing permanent change to its displacement
Fagbenro
In physics, a wave is a disturbance that transfers energy through matter or space, with little or no associated mass transport (Mass transfer). ... There are two main types ofwaves: mechanical and electromagnetic. Mechanicalwaves propagate through a physical matter, whose substance is being deformed
Devansh
K
Manyo
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Doc
Thanks
AMADI
Note: LINEAR MOMENTUM Linear momentum is defined as the product of a system’s mass multiplied by its velocity: size 12{p=mv} {}
AMADI
what is physic
zalmia Reply
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zalmia
Study of matter and energy
Fagbenro
physics is the science of matter and energy and their interactions
Acquah
physics is the technology behind air and matter
Doc
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William
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Easy to learn
William
Practice Key Terms 2

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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