Unformatted text preview: If a force is applied to an object, the object may
experience a change in position, i.e., a displacement.
CHAPTER 6
WORK AND ENERGY
• Work and kinetic energy
o workkinetic energy theorem
• Work done by a variable force
o the dot product When a net force is applied over a
distance, mechanical work is done.
Fx ∆x xo r
F x θ ∆x • Power The work done is W = Fx ∆x ,
• Potential energy
o conservative forces
o nonconservative forces
o nonconservative forces and the workenergy theorem*
o conservative forces and the potentialenergy function*
o equilibrium*
* you study using the handout. where Fx = F cos θ and ∆x = ( x − x o ).
∴ W = ( ma x ) ∆x . 1
2
2
But v 2 = v o + 2a x ∆x ⇒ a x ∆x = ( v 2 − v o ) .
2
1
1
2
∴ W = mv 2 − mv o .
2
2
kinetic energy (K)
This is the workkinetic energy theorem: W = K f − K i = ∆K . Dimensions: F⇒ [M][L]
[ T]2 ∴W ⇒
Units: and ∆x ⇒ [L]. m = 5.0kg F = 80N
∆x [M][L]2
[ T]2 v o = 20m/s (scalar). N ⋅ m ⇒ Joule (J) . Note: work and kinetic energy have the same dimensions
and units. v = 68m/s (a) Use the workenergy theorem: 1
1
21
W = ∆K = mv 2 − mv o = × 5 × ( 682 − 202 )
2
2
2 = 10.6 kJ
(b) Also, W = F.∆x , ∴ ∆x = W 10.6 × 103
=
= 132.5 m.
F
80 Question 1: A constant force of 80N acts on a box of The workkinetic energy theorem mass 5.0kg that is moving in the direction of the force is very important in physics with a speed of 20m/s. A few seconds later the box has
a speed of 68m/s. (a) What is the work done by the In this particular example, F and ∆x are in the same force assuming no friction? (b) How far did the box direction, so the work is positive and the velocity travel in that time? (c) If there had been kinetic friction
between the box and the floor (µ k = 0.5), what would
the final velocity of the box be, in the same time period?
(d) What would be the displacement of the box? increases, because the force F produces a positive
acceleration (a = F ). If F and ∆x are in opposite
m
directions the work would be negative and the velocity
would decrease. So work can be positive or negative. (c) To determine the length of time the force was exerted
(with no friction), we use v = v o + at ⇒ t = v − v o 68 − 20
=
= 3.0s.
80
F
5
m ()() Draw the FBD.
The net force acting on the box is N fk F − f k = F − µ k N = F − µ k mg F
mg = 80 − ( 0.5 × 5.0 × 9.81) = 55.48 N .
So, the “new” final velocity is
F
v = v o + at = v o + net m t = 20 + 55.48 5.0 × 3.0 ( ) = 53.3m/s ( ) theorem:
1
1
F∆x ′ = ∆K = m v 2 − v o2 = × 5 × 53.32 − 202 ,
2
2
i.e., ∆x ′ = 6102 80 = 76.3m.
(Note we use the applied force not the net force.) ) ( 3m from rest by a vertical force of 80N. Find (a) the
work done by the force, (b) the work done by gravity,
and (c) the final velocity of the box. (instead of 68m/s). (d) The displacement is given the workkinetic energy ( Question 2: A box, with mass 6kg, is raised a distance of ) If F is not parallel to the displacement ...
(a) The work done by the
Fy
Fy = 80N 3m r
F applied force Fy is mg 6kg θ = Fy .∆y r
∆s = 80 × 3 = 240J . (b) The work done by the gravitational force is r
r rr
Work done W = ( F cos θ ) ∆ s = F ∆ s cos θ. −( mg).∆y = −(6 × 9.81) × 3 = −177J .
(c) The net work done on the box is ( ) Wnet = Fy − mg .∆y = 240 − 177 = 63J . By the workKE therorem, the net work done is
0
1
1
Wnet = ∆K = mv 2 − mv o2.
2
2
1
2 × 63
∴ mv 2 = 63J ⇒ v 2 =
= 21,
2
6
i.e., v = 4.6 m/s. r
B
θ r
A The dot or scalar product of
r
r
two vectors, A and B is defined as
rr
rr
A • B = A B cos θ.
r
r
If A ⇒ ( A x , A y , A z ) and B ⇒ ( B x , B y , B z )
rr
Then A • B = A xB x + A yB y + A zB z
~ see revision notes on website ~ r
r
Therefore, the work done is W = F • ∆ s r
r
i
j
i
j
We have: A = (3ˆ − 6ˆ ) and B = ( −4ˆ + 2ˆ ).
rr
(a) What is A • B ?
rr
By definition: A • B = A xB x + A yB y
rr
∴ A • B = 3 × ( −4) + ( −6) × 2 = −24 .
What does a negative product imply? rr
Question 3: (a) Find the dot product, A • B, if
r
r
A = (3ˆ − 6ˆ ) and B = ( −4ˆ + 2ˆ ) .
i
j
i
j
r
r
(b) Find the angle between A and B. (b) What is θ ?
rr
rr
rr
A•B
We know A • B = A B cos θ ⇒ cos θ = r r .
AB
r
A = 32 + ( −6)2 = 6.71
r
B = ( −4)2 + (2)2 = 4.47 rr
B
−24
−1 A • r
r
∴ θ = cos
=
A B 6.71 × 4.47 = cos −1 ( −0.80) = 143o. ˆ
j
r
B ˆ
i r
A OK for a constant force. But what
if the applied force is not constant? Consider something simple like catching a ball.
r
FHB r
FBH F
∆x Fx Fx .∆x
x r
∆x Work done W = Fx ∆x = shaded area. Note that the ball applies a force on the hand and the
hand applies an equal and opposite force on the ball,
r
r
i.e., FBH = −FHB.
Assume that in catching the ball the hand moves through
r
a displacement ∆x . Then
r
r
WBH = FBH • ∆x = FBH ∆x ,
r
r
r
r
and WHB = FHB • ∆x = −FBH • ∆x = − WBH .
So work can be positive (work done by the hand ON the
ball in slowing it down) and negative (work done by the
ball ON the hand). Conventionally, when you do work
on a system, we take the work as positive work. x2 x1 F
variable force
Fx i ∆x i x1 x2 Work done over the small displacement ∆x i is Fx i ∆x i . ( ) x2 ∴ Wx1 → x 2 = Limit ∑i Fx i ∆x i = ∫ Fxdx ,
∆x i →0 i.e., area under curve between x1 and x 2. x1 (a) By definition W =
Alternatively, if the force varies in a simple way with
position, we can count squares to find the area.
For example ... x = 3.0
∫ Fx
x =1.5 3.0 = 0.5 ∫ x 3dx =
1.5 Fx (N) dx 0.5 4 3.0
x
= 9.49J .
1.5
4 a (b) Plot the function and count squares. ∫ Fx .dx
0 15 = A1 + A2 + A3. A1 A2 A3 a 0 x( m) A1 = 7.5N ⋅ m, A2 = 25N ⋅ m, A3 = 5N ⋅ m.
∴ Total area = 37.5N ⋅ m Fx ( N) Fx = 0.5x 3 10 5 3 Question 4: A force, Fx = 0.5x N, acts on a object.
What is the work done if the object is moved from x = 1.5m to x = 3.0 m? Solve the problem (a) using
calculus, and (b) graphically. 0 1.0 2.0 3.0 Approximate number of squares:
⇒ 2.5 + 5.75 + 10.5 = 18.75 But 1 square ⇒ 0.5m × 1N = 0.5N ⋅ m
∴ W ≈ 18.8 × 0.5 = 9.4J . x( m) In the previous problem we determined that the amount Interesting cases : (1) lifting a book ... of work done was 46N ⋅ m. Let us see how the work
done is distributed at 2m
intervals by fnding the five F(N)
10
8
6
4
2
0 The net force on the book is:
r
r
r
F − mg = ma y . r
F dy h 0 − 2m : 6 N ⋅ m The incremental work
rr
done is dW = F • dy = Fdy . 2 − 4 m : 10 N ⋅ m So the total work done is smaller areas: x(m)
0 2 4 6 8 10 4 − 6m : 14 N ⋅ m
6 − 8m : 12 N ⋅ m
8 − 10m : 4 N ⋅ m making a total of 46N ⋅ m. Note that positive work is
being done even when the force is decreasing. Negative
work occurs in the following example, when the area is
F(N)
10
8
6
4
2
x(m)
0
0 2 4 6 8 10
−2
−4 negative, which occurs
when F < 0, i.e., when
6 m < x < 10 m. r
mg h h h h W = ∫ Fdy = ∫ ( mg + ma y )dy = mg ∫ dy + m ∫ a ydy .
0 0
h 0
h dv But, m ∫ a ydy = m ∫
0 0 dt 0 v v
dy
dv = m ∫ vdv
v o dt
vo dy = m ∫ v
1
= mv 2 = 0, if v o = v ( = 0) ,
2
v o i.e., the book starts and finishes at rest (so ∆K = 0 ).
h ∴ W = mg ∫ dy = mgh .
0 Therefore, the work done against the gravitational force,
does not depend on how the fast/slow the book is lifted,
but only on the height! Interesting cases: (2) a block sliding down a slope ...
N l y Power is the rate at which work is
done (i.e., quickly or slowly). For x h POWER example, climbing a flight of stairs θ θ
mg quickly requires more power than climbing the same mg flight of stairs slowly. As the block slides down the slope, with no friction, the
only force on the block in the xdirection is due to the
gravitational force: At any instant (i.e., instantaneous power)
rr
dW F • ds r r
P=
=
= F • v (scalar).
dt
dt Fx = mg sin θ The work done by the gravitational force on block is:
W = Fx .l = mgl sin θ . Dimensions: Power ⇒ But l sin θ ⇒ h the height of the slope. ∴ W = mgh .
Note that W is independent of angle θ; the work done by
gravity depends only on the vertical height not the
length of the actual path! = work done [M][L]2
⇒
time
[ T]2 [ T] [M][L]2
.
[ T]3 Units: J/s ⇒ watts (W)
1000W ⇒ 1kW
1HP ⇒ 746W Energy and work are one thing ... but power is quite
another.
Look ... one gallon of gas has a Onedimensional example: certain energy content and so can
do a fixed amount of work ~ 120 × 106 J . rr
P = F • v = Fx v x = ma x v x
P
i.e., a x =
.
mv x If the engine operates with constant power output, the
But the power produced when the gas burns can have any resulting acceleration is inversely proportional to the value ... it depends on how fast it burns! velocity, i.e., as v increases, a decreases. Also
dv
P = Fx v x = ma x v x = mv x x
dt A gallon of gas will provide enough
power to operate a lawnmower for = hours (a few HP). d 1
dK
mv x2 =
, dt
dt 2 i.e., the instantaneous power is the rate of change of
However, a gallon of gas may only
provide enough power to operate a
jet engine for ~ 1 minute!
(~ 2500HP). kinetic energy. We have just found that P = dK
.
dt 1
∴ P.dt = dK = d mv 2 = mv.dv .
2 If the power and mass are constant, and the car starts at
rest at t = 0 , then
t
v
1
2Pt
P ∫ dt = m ∫ v.dv ⇒ Pt = mv 2 ⇒ v =
,
2
m
0
0 i.e., v ∝ t1 2 and a = dv
∝ t −1 2 .
dt Question 5: A electric motor supplies constant power
(P) to the wheels of car of mass (M). If the car starts v from rest, show that its velocity after a time t along a v∝ t horizontal road is
v= ignoring friction. 2Pt
,
M t a
a∝1 t t Conversion factors: 1HP = 746 W ⇒ 10HP = 7.46 kW .
1mi/h = 0.45m/s ⇒ 10 mi/h = 4.5m/s,
⇒ 20 mi/h = 9.0 m/s,
⇒ 30 mi/h = 13.5m/s, Question 6: An electric vehicle, of mass 500kg, is
powered by a motor that has a constant power output of
10HP. If all of the power of the motor produces motion, ⇒ 40 mi/h = 18.0 m/s.
∆K
∆K m
=
v f 2 − v i2
From earlier P =
. ∴ ∆t =
∆t
P
2P ( = 500
2
2
2
2
3 v f − v i = 0.0335 v f − v i
2 × 7.46 × 10 ( ) find the time taken for the vehicle to accelerate (a) from
0 to 10mi/h , (b) from 10 to 20mi/h, (c) from 20 to
30mi/h, and (c) from 30 to 40mi/h. ) () (a) ∆t = 0.0335 4.52 = 0.68s. ( ) (b) ∆t = 0.0335 9.02 − 4.52 = 2.04s. ( ) (c) ∆t = 0.0335 13.52 − 9.02 = 3.39s. ( ) (d) ∆t = 0.0335 18.02 − 13.52 = 4.75s. ( ) In the previous problem we found
2Pt
v=
= 5.463 t ,
m
using the values of P and m given here.
v ( m/s)
40mi/h 20 ×
∆v 16 30mi/h
20mi/h
10mi/h 12 ×
8
3.39s
∆v
×
4 2.04s
0 2 4 ×
∆v Question 7: A 5kg box is being lifted vertically at a 4.75s constant speed of 2 m/s by a force equal to the weight of
t(s) 6 8 10 12 14 Notice that the time taken to increase speed by 10mi/h
increases as the initial speed increases since
m
∆t =
v 2 − v i2 ∝ ( v f − v i )( v f + v i ),
2P f ( ) where ( v f − v i ) = ∆v = 10mi/h .
∴ ∆t ∝ ( v f + v i ) ,
since ∆v is constant. the box. (a) What is the power input of the force? (b)
How much work is done by the force in 4s? Note, since there is no net force acting on the box, it is
traveling at constant speed.
Fy ( = mg) (a) Fy = mg = 5 × 9.81 = 49.1N.
Therefore, the power supplied by mg force Fy is
rr
PF = F • v = Fy v y = 49.1 × 2 = 98.1W POTENTIAL ENERGY
So far we have seen that when work is done on a isolated
object, it leads to a change in kinetic energy. However,
there are systems on which work does not lead to a
change in kinetic energy. We will look at two examples.
1. Springs: consider a spring inside a toy gun. (b) By definition, power is the rate at which work is
done, i.e., P = ∆W ∆T .
Therefore, the work done by Fy in 4s is r
F1 r
F2 r
x1 r
x2 ∆W = P.∆t = 98.1 × 4 = 392J . Note that the power associated with the gravitational
force is Pg = − mgv y = −49.1 × 2 = −98.1W .
So, the net power supplied is (PF − Pg ) = 0.
Since P = ∆K ∆t , ∆K ⇒ constant,
i.e., there is no change in kinetic so the box is traveling at
constant speed (which we already knew!). You compress the spring by applying two equal and
opposite forces F1 and F2. The net force on the spring is
zero and so there is no change in the kinetic energy of the
spring. However, you have done work on the spring
r
r
r
r
W = F1 • ∆x1 + F2 • ∆x 2 = 2F∆x ,
r
r
r
r
since F1 = F2 = F and ∆x1 = ∆x 2 = ∆x . r
F1 r
F2
r
x1 r
x2 2
1 2
h 1 h mg Wg = − mgh Wyou = mgh So, what happened to the work you did on the spring?
Clearly, the configuration of the spring has changed as In lifting an object from 1 to 2 through a distance h, evidenced by the change in length. In fact, the spring has the work done by you on the system is Wyou = mgh . But stored the work you did as potential energy, and when
released it will transfer the stored potential energy to the
ball. the object starts and finishes at rest so ∆K = 0 . The
object is acted on by two equal and opposite forces  the
force you apply upward ( mg) and the force the Earth 2. Raising an object in the Earth’s gravitational field .
In this scenario we must remember that the ball and the
h Earth are a system; so this is a two applies downward ( − mg). So the net work done is
Wnet = Wyou + Wg = mgh + ( − mgh ) = 0, ∴ ∆K = 0 .
What happened to the work you did on the system? We particle system. However, you are not note that at 2 the object is capable of doing work; for part of the system; you are an external example, when released it would fall so it could strike a agent that does work on the system. nail and drive it into the floor! The amount of work the object can do is mgh , which is
the same as the amount of work
you, an external agent, did raising
v path is one where the displacement is zero. In fact, at 2 the object/Earth h on an object around any closed path is zero ... a closed the object from 1 to 2 . 2 A force is called conservative if the total work it does Note: what the work W2→1 does ...
... it increases the velocity of the object 1 ( ) system stores this work as potential energy Ug = mgh ,
i.e., the system has the potential to return the work the W2→1 = ∆K = K1 − K2 = K1.
Thus, the work you did on the object/Earth system (mgh) external agent (you) did, when the object is released. has been converted into kinetic energy of the object. Note: when the object is released (from rest)
1
∆K = W = mgh = mv 2,
2 Complementary definition: so its velocity at impact is v = 2 gh . The work done by a conservative force on an object is
zero when the particle returns to its initial position, i.e., When the object is released, the work done by the
gravitational force on the object is: rr
Mathematically: W ≡ ∫c F • ds = 0 , where ∫c denotes an W2→1 = + mgh . Thus, the net work done by the gravitational force is
1 − mgh 2 mgh when the particle moves around any closed path. 1 = 0. integral around a closed path. Conservative forces ...
What about nonconservative forces ? 2 z b
a
y
1 x If the work done in going from 1 ⇒ 2 is:
W1→2 then (by definition) the work done in going from 2 ⇒ 1
with a conservative force is
W2→1 = − W1→2 along any path. The work done by a conservative force
along any path is determined only by the end points not Move a book across a table from A to B with kinetic
friction; you do work against the frictional force, i.e.,
WA→ B = Fk .d A→ B = mgµ kd A→ B.
Consider two paths a and b. Since b > a , then Wb > Wa . So the work done against the frictional force does depend
on the path!
~ The frictional force is a non conservative force ~
** What happens to the extra work ? ** the route.
NOTE: The work done by a nonconservative force is
• The gravitational force is a conservative force.
• The elastic force is a conservative force. nonrecoverable (it produces heat, sound, etc.) r
F y
(0, l) r
F=0 r
dl (l, l) r
dl
r
dl r
F r
F = clˆ
j Question 8: In a certain region of space, the force on an
electron is (0, 0) r
ˆ
F = cxj, where c > 0. The electron moves in a counterclockwise r
dl ( x, y ) = (0, 0),(l, 0),(l, l),(0, l), ˆ
i x The work done along each leg of the loop can be
calculated from the expression direction around a square loop in the xy − plane. If the
corners of the loop are at (l, 0) ˆ
j W1→2 = r
F • dl . x2 ,y2 r ∫ x1 , y1 one complete trip around the loop. Is the force a Along the leg from (0, 0) to (l, 0) the force increases
rr
linearly with x, but at all points F⊥dl , so the dot product conservative or nonconservative force? is zero. Similarly for the leg from (l, l) to (0, l). Thus, find the work done on the electron by the force during no work is done on the electron as it travels along the
legs parallel to the xdirection, i.e., W1→2 = W3→ 4 = 0.
Along the leg from (l, 0) to (l, l) the force is
r
F = clˆ .
j r
F y
(0, l) r
dl r
F=0 r
dl r
F (0, 0) Nonconservative forces and the workenergy theorem (l, l) r
dl r
F = clˆ
j (l, 0) r
dl ˆ
j ˆ
i Consider an object falling with airresistance. There are
two forces to consider; the gravitational force
(conservative) and the drag force (nonconservative). x 1 W1→2 = Wc + Wnc , Thus, the work done along that leg is
lr
l
r lr
ˆ
W2→3 = ∫ F • dl = ∫ F • dyj = cl ∫ dy = cl2 .
0 0 The total work done is
where Wc is the work done by the
conservative (gravitational) force and 2 0 Wnc is the work done by the non Along the leg from (0, l) to (0, 0) the force is zero and so conservative (drag) force. But, by the workenergy no work is done on the electron, i.e., W4→1 = 0. Thus,
the total work done on the electron for the round trip, theorem starting at (0, 0), is W = W1→2 + W2→3 + W3→ 4 + W4→1 = cl2 . W1→2 = K2 − K1.
Also Wc = −( U2 − U1 ). ∴ K2 − K1 = −( U2 − U1 ) + Wnc ,
i.e., Wnc = (K2 + U2 ) − (K1 + U1 ). The start and end points are the same but the total work
r
done by the force is nonzero. Therefore, F is a non Therefore, the work done by a nonconservative force is conservative force. Since W > 0, the energy of the equal to the change in mechanical energy. electron increases as it goes around the loop. Conservative forces and the potential energy function
Given that Fx = −
If the potential energy of a system has a unique value at
r
every point r , we can define a potential energy function,
r
U( r ). Therefore, the potential energy function tells us
how potential energy varies with position.
One property of a conservative force is that the work
done by the force can be expressed as the difference
between the initial and final values of the potential
energy, i.e., dU
, if we know the potential energy
dx function, U, we can determine the force Fx at any point.
Conversely, if we know the functional form of Fx, we can
find the potential energy function, since dU = −Fx .dx ⇒ U = − ∫ Fx .dx .
This is true, generally. Examples ...
[1] Gravitational force:
ˆ
k rr
dW = F • ds = −dU. Let us find the potential energy
r
mg Hence, the work done by a conservative force equals the
decrease in the potential energy function from point 1 to function. We take the zdirection
ˆ
( k ) vertical. Then, from above, the incremental change in potential energy as the object falls point 2.
2r r
∴ ∆U = U2 − U1 = − ∫ F • ds .
1 rr
Since dU = −F • ds , then, in onedimension,
dU
dU = −Fx .dx , i.e., Fx = − .
dx is rr
ˆ
ˆ
ˆ
ˆ
dU = −F • ds = −( − mgk ) • (dxi + dyj + dzk ) = ( mg)dz . ∴ U = mg ∫ dz = mgz + Uo,
where Uo is the reference energy when z = 0 . Let’s plot
this function: Note: U ⇒ U( z), i.e., it is
U(z) uniquely defined at z. This 2 ∆U = U2 − U1 Uo is the potential energy Height (z) 1 Therefore, the elastic potential energy function is:
rr
dU = −F • ds = −( −F2 ).dx = kx.dx . function for the
gravitational force. 1
∴ U = k ∫ x.dx = kx 2 + Uo .
2 But, when there is no displacement, i.e., x = 0, the spring Normally, we deal only with differences in potential has zero elastic potential energy. Therefore, U(0) = 0, energy, so the choice of Uo is entirely arbitrary. Note: the and so Uo = 0 . Also, at this point, there is no net force force associated with this potential energy function is
dU
Fz = −
= − mg, as expected.
dz acting on the spring (in the xdirection); we say that the
spring is in equilibrium. This is the definition of
equilibrium, which we first came across in chapter 4
(Newton’s 1st Law). [2] Elastic force:
If you compress a spring F1 = kx Equilibrium a distance x in the xF2 = − kx
x direction, the force, F1,
you exert is given by
Hooke’s Law, i.e., At equilibrium we have F = 0, i.e., dU
= 0, so U(x) is an
dx does negative work since the force it exerts on you extremum. However, we must exercise care in using that
dU
= 0 does not uniquely define a
criterion, since
dx (F2 = − kx ) is opposite to the displacement x. minimum in the potential energy function. F1 = kx , where k is the spring constant. But the spring U(x) Equilibrium position: x o.
Stable equilibrium when
U( x o ) is a minimum,
x i.e., d2U
> 0.
dx 2 xo
U(x) Equilibrium position: x o.
Unstable equilibrium when
U( x o ) is a maximum,
x d2U
< 0.
i.e.,
dx 2 xo Question 9: The potential energy of an object, confined
to move along the xaxis, is U = 3x 2 − 2 x 3 for x ≤ 3m,
and U = 0 for x > 3m,
where U is in Joules and x is in meters. If the only force
acting on the object is the force associated with this
potential energy function, (a) at what positions is the
object in equilibrium and what type of equilibrium is U ( x) Equilibrium position: x o. associated with each position? (b) Sketch the potential Neutral equilibrium since energy function. U( x o ) is an inflexion point, i.e.,
x xo d2U
= 0.
dx 2 (a) To find the equilibrium positions we need F( x ).
dU
F( x ) = −
= −6 x + 6 x 2 .
dx
F(x) At equilibrium F( x ) = 0 ,
dU
i.e.,
= 6x2 − 6x
dx = 6 x ( x − 1) = 0 . x( m)
−1 1 2 (b) Therefore, the equilibrium U( J ) positions are at x = 0 and x = 1. 10 U( x ) = 3x 2 − 2 x 3 To find out the type of equilibrium look at the signs of the
2 seond derivative, dU
, at x = 0 and x = 1.
dx 2
2 Now, dU
= 6 − 12 x .
dx 2 ∴ at x = 0 : d2U
= 6 (> 0) ⇒ Stable.
dx 2 ∴ at x = 1: d2U
= −6 ( < 0) ⇒ Unstable.
dx 2 x(m) −1 1
−10 2 m y
N Fs k x •
fs m k mg θ 1
The potential energy of the spring is U = kx 2 and so we
2 θ need to determine the extension of the spring produced by Question 10: A block of mass m rests on an inclined the force just necessary to start the block sliding. From plane, as shown above, and the coefficient of static the freebodydiagram, just as the block slides, we have:
∑ Fx = Fs − f s − mg sin θ = 0, friction between the block and plane is µ s. A gradually
increasing downward force is applied to the spring, of
force constant k, until the block just starts to move.
Show that the potential energy of the spring at that point
is
U= [ mg(sin θ + µs cos θ)]2 .
2k and ∑ Fy = N − mg cos θ = 0.
But f s = µ s N = µ s mg cos θ and Fs = kx .
∴ kx − µ s mg cos θ − mg sin θ = 0,
mg(sin θ + µ s cos θ)
i.e., x =
.
k
2 1
[ mg(sin θ + µs cos θ)] .
∴ U = kx 2 =
2
2k (QED) ...
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Full Document
 Spring '08
 Guzman
 Physics, Energy, Force, Kinetic Energy, Potential Energy, Work, mgh

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