Up to this point, you've been solving problems for which
all the forces were conservative. By taking your system to include all
objects interacting through conservative forces, you were able to say
mechanical energy was conserved. In that case, the general conservation of energy
equation W_{ext}_{ }= ΔE_{sys} reduced to 0 =
ΔE_{sys}. In the event that
there are forces external to the system that do work, the lefthand side will
no longer be 0. Any force that is external to the system and does work
on the system will be represented in W_{ext}.
Here's a situation with an external force that does work on
the system. Open this animation. A block is initially given a push to start it
moving upward between two rails. The rails exert kinetic friction force on
the block. We take the system to include the block and the Earth. Kinetic
friction is an external force that does work on the system. Note that the U_{g} = 0 level is set at the starting position. Also note the following.

At t = 0 (Figure 1), all of the system energy is
kinetic.

Figure 2 shows that as the block rises and slows
down, U_{g} increases while K decreases. The system energy,
however, isn't conserved. That's because the work done by the friction
force on the block is negative, since the friction force is always opposite the
displacement. Note also that the change in E_{sys}, which is
negative, is equal to W_{f}. Therefore, W_{ext} = W_{f}_{ }= ΔE_{sys} as expected.

At the highest position (Figure
3), the kinetic
energy is 0, since the block stops momentarily. In this situation, all
of the system energy is gravitational potential. However, the system
energy has continued to decrease while the work done by friction has become
more negative.

When the block returns to its starting position
(Figure 4), the gravitational potential energy has returned to 0. This
is because the net displacement in the round trip is 0. The work done by friction,
however, continued to become more negative even after the block reversed
directions. This is because friction is not a conservative force.
Figure 1 
Figure 2 
Figure 3 
Figure 4 




t = 0 
partway up 
highest position 
back to the starting position 
Now let's look at some workedout examples.
Example 1. A 17g leaf falls
from a tree to the ground. If the leaf was 5.30 m
above the ground as it left the tree and its speed just before landing was 1.3 m/s, how work did air
friction do on the leaf?
Given: v_{i} = 0 v_{f} = 1.3 m/s y_{i} = 5.3 m y_{f} = 0 m = 0.017 kg
Goal: Find the work done by air friction on the
leaf.
System: leaf, Earth
External forces: air friction
initial state: leaf detaches from tree
final state: leaf reaches ground
Energy changes:
ΔU_{g} is negative
ΔK is positive 
 The direction of +y is defined to be up.
 The lowest position of the leaf is designated as y_{f}= 0.



Write the general
conservation of energy law. 
W_{ext} is W_{air},
because the force of air friction does work on the leaf. The energy changes in the system are
kinetic and gravitational. 
Expand the delta notation in terms of
initial and final quantities. 
The initial kinetic energy and final
gravitational energy are 0. 
Substitute expressions for the
remaining energy terms. 
Substitute the givens with units. 
The work done by air friction is
negative. This is because the force of air friction on the leaf is
opposite the direction of the leaf's displacement. 
Example 2. The problem is 834, part a,
on page 235 of the text. A 42.0kg seal at an amusement park slides
from rest down a ramp into the pool below. The top of the ramp is 1.75 m
higher than the surface of the water and the ramp is inclined at an angle of
35.0° above the horizontal. If the seal reaches the water with a speed of
4.40 m/s, what is the work done by kinetic friction?
Given: v_{i} = 0 v_{f} = 4.40 m/s y_{i} = 1.75 m y_{f} = 0 θ = 35° m = 42.0 kg
Goal: Find the work done by kinetic friction.
System: seal, Earth
External forces: normal, kinetic friction
initial state: highest position of seal
final state: lowest position of seal
Energy changes:
ΔU_{g} is negative since the
seal decreases in elevation;
ΔK is positive since the seal speeds up 
 The direction of +y is defined to be up.
 The lowest position of the seal is designated as y_{f}= 0.

. 

Write the general
conservation of energy law. 
W_{ext} = W_{N} + W_{f},
because the normal force and kinetic friction are both external forces.
However W_{N} = 0, because the normal force is
perpendicular to the displacement. 
The energy changes in the system are
kinetic and gravitational. 
Expand the delta notation in terms of
initial and final quantities. 
The initial kinetic energy and final
gravitational energy are 0. 
Substitute expressions for the
remaining energy terms. 
Substitute the givens with units. 
Note the work done by friction is
negative as expected. 
Example 3. A 64kg jogger is
running at a speed of 1.1 m/s as he reaches the bottom of a hill. He runs up
the hill and has a speed of 3.5 m/s at the top of the hill. Air friction did
a total amount of work of 3500 J on the jogger while he was running up the
hill. The hill is 21 m high. How much work did the ground do on the jogger
as he was climbing the hill?
Given: v_{i} = 1.1 m/s v_{f} = 3.5 m/s y_{i} = 0 m y_{f} = 21 m m = 64 kg W_{f} = 3500 J
Goal: Find the work done by the ground on the jogger.
System: hill, Earth
External forces: normal,
air friction,
ground pushing jogger forward
initial state: lowest position of jogger
final state: highest position of jogger
Energy changes:
ΔU_{g} is positive since the
jogger increases in elevation;
ΔK is positive since the jogger speeds up 
 The direction of +y is defined to be up.
 The lowest position of the jogger is designated as y_{i}= 0.

. 

Write the general
conservation of energy law. 
W_{ext} is composed of
three terms. The normal force does 0 work on the jogger. Air friction
does negative work on the jogger. By
Newton's 3rd law, the force of the jogger's feet backward on the ground
is equal and opposite to the force of the ground forward on the jogger's
feet.
The latter force does positive work on the jogger. 
The energy changes in the system are
kinetic and gravitational. 
Expand the delta notation in terms of
initial and final quantities. 
The initial gravitational energy is 0. 
Solve for the work done by the
ground. 
Substitute the givens with units. 
Note the work done by the ground is
positive as expected. 
Example 4. A block of mass 1.8 kg is connected to a second
block of mass 1.2 kg as shown in the diagram below. When the blocks are
released from rest, they move through a distance of 0.30 m before the second
block hits the floor. The coefficient of kinetic friction between block 1
and the table is 0.35. Find the speed of either block just before block 2
hits the floor. Ignore the mass of the pulley and the string, and assume
that the string doesn't stretch.
Given: v_{i} = 0 m/s y_{i,2} =
0.30 m y_{f,2} = 0 m m_{1} = 1.8 kg m_{2} =
1.2 kg μ_{k } = 0.35
Goal: Find the speed v_{f} of either
block before block 2 hits the floor.
System: both blocks, Earth
External forces:
normal on block 1, tension on both blocks by the string
initial state: both blocks at rest
final state: both blocks moving at the same speed just before block 2
hits the floor
Energy changes:
ΔU_{g} is negative since block 2 decreases in elevation;
ΔK is positive since both blocks speed up 
 The final state is shown by the dashed
lines below.
 The direction of +y is defined to be up.
 The lowest position of the block 2 is designated as y_{f}= 0.

. 

W_{ext} is composed of
three terms. The normal force does 0 work on block 1, while kinetic
friction does negative work. Together, the two tension forces do 0 work
on the system. The tension force pulling to the right on block 1 does
positive work, since the force is in the same direction as the
displacement. The tension force pulling up on block 2 does negative
work, since the force is opposite the displacement: W_{T} = Tdcos0° + Tdcos180° = 0. 
The energy changes in the system are
kinetic and gravitational. We expand these in terms of the changes for
each block. The kinetic energy changes can be grouped together, since
the blocks move together. We keep the potential energy changes separate.
Block 1 experiences no change in gravitational potential energy, since
the elevation of that block doesn't change. Block 2 experiences a
decrease in gravitational potential energy. 
The work done by kinetic friction on
block 1 is W_{f} = μ_{k}NΔx(cosθ) = μ_{k}M_{1}gΔx(cos180°)
= μ_{k}M_{1}gΔx. Note that the
horizontal displacement Δx of block 1 is the negative of the vertical displacement Δy = (y_{2f}  y_{2i})
of block 2. Since y_{2f}= 0, Δx = y_{2i}. 
We select the positive root since we are asked for speed. 
Example 5. See this animation. At t = 0, a block attached to a spring is
released from rest and oscillates horizontally about the origin on a table.
There is friction between the block and the table. How much work does
friction do on the block from t = 0 until the block returns for the
first time to its turnaround point on the right? The mass of the block is
1.5 kg, and the spring constant of the spring is 460 N/m.
Given:
x_{i} = 0.090 m
x_{f} =
0.064 m
v_{i} = 0 m/s
v_{f} = 0 m/s
m = 1.5 kg
Goal: Find the work done by kinetic friction on the
block from t = 0 to the first turnaround point on the right.
System: block, spring
External forces: normal, gravity, kinetic friction
initial state: block at rest at x_{i}
final state: block momentarily at rest at x_{f}
Energy changes:
ΔU_{el} is negative since the final extension of the
spring is less than the initial extension.
ΔK is 0. 
 x = 0 is the equilibrium position
of the spring.
 The initial and final positions are read
from the animation.

. 

W_{ext} is composed of
three terms. The normal and gravitational forces do 0 work on the block, while kinetic
friction does negative work. 
The only energy change in the system
is elastic. The initial and final velocities are 0, so the change in
kinetic energy is 0. The elastic potential energy decreases. 
In finding the change in elastic
potential energy, find the differences of the squares of the two
extensions rather than finding the square of the difference. These two
operations aren't equivalent. 
Note that it's not necessary to know
the mass of the block to solve the problem. 
The work done by friction is negative as expected. 
