Energy Problem examples and lecture notes for Monday, October 14, 2019

We’ve looked at the energy in a system as being a conserved quantity. If there are no external

forces that act on your system then mechanical energy is conserved, as there is nothing external to the

system to do work on the system. The energy of the system therefore does not change. We write this

as ΔE = ΔK + ΔUg + ΔUs = 0 , where ΔK = ΔK1 + ΔK2 + ... is the change in kinetic energy for each of

the objects in the system that is moving, ΔUg = ΔUg1 +Ug 2 + ... is the change in the gravitational

potential energy of each of the objects in the system due to changing the objects height, and

ΔUs = ΔUs1 +Us2 + ... is the change in elastic potential energy in each of the springs in the system.

If there is friction in the system then we consider it to be an external force. Friction takes energy

out of the system and mechanical energy is not conserved and there is a change in energy of the

system. To incorporate friction change in energy statement we say the change in energy is due to the

work done by friction. We write work done by friction as ΔE fr =Wfr = FfrΔxcos 180( ) = −FfrΔx , and

thus ΔE fr = −FfrΔx = K + ΔUg + ΔUs .

Example 1: Two masses connected to a spring from class on Friday.

Consider the arrangement of masses shown below connected to a massless pulley. Mass

m1 = 10kg is connected to a second mass m2 = 2kg . Mass m2 is also connected to a massless spring with stiffness k = 1000 N

m . If the system is released from rest, by how much does the spring extend from its starting (or equilibrium) position?

m1

m2

k

Solution:

Take the origin of the system to be at m2 . Mass m2 will rise by a height y , extending the spring by an amount y while also allowing mass m1 to fall through a height y . If we take the system as the masses, spring and the world then there is nothing external to the system and energy is conserved. Applying conservation of energy and noting both masses start from rest and when the spring is fully extended both masses return to rest we have

ΔE = 0 = ΔK + ΔUg + ΔUs = ΔK1 + ΔUg1( ) + ΔK2 + ΔUg 2( ) + ΔUs

0 = ΔUg1 + ΔUg 2 + ΔUs = m1gy1 f − m1gy1i( ) + m2gy2 f − m2gy2i( ) + 12 ky f

2 − 12 kyi

2( )0 = 0− m1gy( ) + m2gy − 0( ) + 1

2 ky2 − 0( ) = y −m1g + m2g + 12 ky( )

∴ y =0

2 m1 − m2( )gk

⎧

⎨⎪

⎩⎪

y =2 m1 − m2( )g

k=

2 10kg − 2kg( )9.8 ms2

1000 Nm

= 0.16m

The spring is extended from its equilibrium position by 0.16m = 16cm . As a check on the signs in our statement of conservation of energy we have the potential energy in the stretched spring increasing (positive), mass m1 falling, so it loses gravitational potential energy (negative) while the gravitational potential energy of mass m1 increases as it rises (positive). All of the loss in gravitational potential energy in mass m1 contributes to the increases in potential energy of mass

m2 and stretching the spring. This can be seen in line #3 above. Example 2: Falling masses on the pulley

Consider the system in example 1. We found that the spring is extended by an amount y = 16cm when the masses return to rest. How fast is mass m1 moving when, say the spring is extended by an amount y = 10cm from its equilibrium position, assuming that the system of masses is released from rest?

Solution:

Again, we apply conservation of energy with both masses starting from rest. We have

ΔE = 0 = ΔK + ΔUg + ΔUs = ΔK1 + ΔUg1( ) + ΔK2 + ΔUg 2( ) + ΔUs

0 = 12 m1v1 f

2 − 12 m1v1i

2( ) + m1gy1 f − m1gy1i( ) + 12 m2v2 f

2 − 12 m2v2i

2( ) + m2gy2 f − m2gy2i( ) + 12 ky f

2 − 12 kyi

2( )0 = 1

2 m1v f2 − 0( ) + 0− m1gy( ) + 1

2 m2v f2 − 0( ) + m2gy − 0( ) + 1

2 ky2 − 0( )12 m1 + m2( )v f

2 = m1 − m2( )gy − 12 ky2

v f =2 m1 − m2( )gy − ky2

m1 + m2( ) =2 10kg − 2kg( )9.8 m

s2 × 0.1m−1000 Nm 0.1m( )2

10kg + 2kg( ) = 0.47 ms

Here we note that since the cord connects both masses they have the same final speeds. As a check on the signs in our statement of conservation of energy we have the kinetic energies of both masses increases (both are positive) as well as the potential energy in the stretched spring increasing (positive). Mass m1 falls, so it loses gravitational potential energy (negative) while the gravitational potential energy of mass m1 increases as it rises (positive). This can be seen in line #3 above.

Example 3: Pinball

In a pinball game with marbles, a 10 Nm spring is compressed 3cm releasing a 50g

marble from rest. If the marble needs to travel 60cm up a 30 incline before entering the scoring zone of the game table, will it make it? If not, how much must the spring be compressed so that it will?

Solution: The initial energy stored in the spring as potential energy is converted to a gravitational

potential energy as the marble rises vertically and travels a distance d along the ramp. To determine the distance d we use conservation of energy with the final speed of the marble equal to zero. If the distance traveled by the marble before coming to rest is greater than 60cm the marble makes it into the scoring zone and if the distance is less than 60cm it does not make it into the scoring zone. We have

ΔK + ΔUs + ΔUg = ΔE = 012 kx f

2 − 12 kxi

2( ) + mgy f − mgyi( ) = − 12 kxi

2 + mgy f = − 12 kxi

2 + mgd sinθ = 0

d =kxi

2

2mg sinθ=

10 Nm × 0.03m( )2

2× 0.050kg × 9.8 ms2 × sin3

= 0.175m = 17.5cm

.

Since this is less than the 60cm the ball does not make it. In order to have the ball travel 60cm we need to compress the spring by an amount given by

ΔK + ΔUs + ΔUg = ΔE = 012 kx f

2 − 12 kxi

2( ) + mgy f − mgyi( ) = − 12 kxi

2 + mgy f = − 12 kxi

2 + mgd sinθ = 0

xi =2mgd sinθ

k=

2× 0.050kg × 9.8 ms2 × 0.60m× sin3

10 Nm

= 0.056m = 5.6cm

Example 4: Roller Coaster

A roller coaster car, with a mass of 500kg , crests a 20m high hill while moving at a speed of

10 ms . It then rolls down the other side, all the way to ground level, before climbing a second

hill.

a . What is the speed of the car when it is 10m up the second hill? b. What is the maximum possible height of the second hill?

c. If the car is subject to a frictional force that causes it to lose 8000J of energy, what is the

maximum height of the second hill? Solution:

a.

To find the speed, we use conservation of energy and we have,

ΔK + ΔUs + ΔUg = ΔE = 012 mv f

2 − 12 mvi

2( ) + mgy f − mgyi( ) = 0

v f = vi2 + 2g yi − y f( ) = 10 m

s( )2+ 2× 9.8 m

s2 × 20m−10m( ) = 17.2 ms

b. The maximum height of the second hill is determined when the cart stops. Again, using

conservation of energy ideas, we have

ΔK + ΔUs + ΔUg = ΔE = 012 mv f

2 − 12 mvi

2( ) + mgy f − mgyi( ) = 0

y f =vi

2

2g+ yi =

10 ms( )2

2× 9.8 ms2

+ 20m = 25.1m

c. Here friction is an external force and friction changes the energy of the system, by taking energy

out of the system. The change in energy of the system, due to friction, using energy ideas is

ΔK + ΔUs + ΔUg = ΔE fr =Wfr

12 mv f

2 − 12 mvi

2( ) + mgy f − mgyi( ) = −8000J

y f =vi

2

2g+ yi −

8000Jmg

=10 m

s( )2

2× 9.8 ms2

+ 20m− 8000J500kg × 9.8 m

s2

= 23.5m

Example 5: Springs on a horizontal surface

Block #1 is attached to a horizontal spring and slides on a frictionless horizontal surface. Block #2 has the same mass as #1 and also sits on the same frictionless surface. It is attached to a spring with three times the stiffness of the other one. If both blocks have the same amplitude of motion find the ratio of the following quantities (#2/#1): the periods of the motion, the angular frequencies, the maximum velocities, the maximum accelerations, and the maximum displacements.

Solution:

a. The period of oscillation of the mass is given by T = 2 mk

π so, with identical masses, taking a

ratio gives

2 1

1 2

T kT k

= , but k2 = 3k1, so 2

1

13

TT

= = 0.58.

b. The angular frequency is given by km

ω = , so again taking ratios, 2 2

1 1

3kk

ωω

= = = 1.73

c. The maximum velocity is given by vmax =ω xmax , so with the same xmax we have

max,2 2

max,1 1

1.73vv

ωω

= =

d. The maximum acceleration is given by amax =ω2xmax , so

2max,2 2 2

2max,1 1 1

3a ka k

ωω

= = =

Example 6: Another mass on a spring

Attached to a spring on a frictionless tabletop, a 1kg mass is observed to undergo horizontal simple harmonic motion (it oscillates back and forth) with a period of 2.5s after stretching the spring. The spring is then held vertically and a 0.2kg mass is attached and gently lowered to its equilibrium position.

a. Find the distance the spring is stretched.

b. If the spring is then stretched an additional 5cm and released, find the period of the

subsequent motion.

c. What is the maximum acceleration of the 0.2kg mass?

d. What is its maximum velocity of the mass? Solution:

a. We calculate the stretch from Hooke’s law. To do this we need the spring constant, which we get from the information on the horizontal mass and the period.

( ) mN

skg

Tmk

kmT 32.6

5.21442 2

2

2

2

=×==→= πππ and from Hooke’s law,

cmmkg

kmg

kFxxkF

mN

sm

3131.032.6

8.92.0 2 ==×

===Δ→Δ=

b. The period of the oscillating mass on the vertically oriented spring is given by

skgm

kT mN

12.12.032.6

22 === ππ .

c. The maximum acceleration of the mass is given by

amax =ω

2xmax =km

xmax =6.32 N

m

0.2kg× 0.05m = 1.58 m

s2 .

d. The maximum velocity of the mass is given by

vmax =ω xmax =

km

xmax =6.32 N

m

0.2kg× 0.05m = 0.28 m

s

Example 7: An amusement park thrill ride

An amusement park thrill ride consists of a cart with some riders (of total mass 500kg ) that is set in motion by a large spring with spring constant 800 N

m . The cart travels along the flat horizontal section of track that is located 15m above the ground and then down the ramp toward the loop-the-loop, which has an unknown diameter, D . The entire track is frictionless and is shown in the figure below.

a. If the spring is initially compressed by 3m , what is the speed of the cart as it leaves the spring?

b. What is the speed of the cart at the bottom of the ramp before the loop-the-loop?

c. If the speed of the cart at the top of the loop-the-loop is 8.55 m

s , what is the diameter of the loop?

d. How much work was done by gravity on the cart as it traveled from the bottom of the loop-

the-loop to the top?

e. Suppose that the horizontal section of the track at the top were not frictionless, but that a frictional force was present, with a coefficient of kinetic friction of µ = 0.2 . What would the speed of the cart be as it left the spring?

Solution:

a. Applying conservation of energy we have

ΔE = ΔK + ΔUg + ΔUs = 0

0 = 12 mv f

2 − 12 mvi

2( ) + mgy f − mgyi( ) + 12 kx f

2 − 12 kxi

2( ) = 12 mv f

2 − 12 kxi

2

∴ 12 kxi

2 = 12 mv f

2 → v f =km

xi2 =

800 Nm

500kg−3.0m( )2

= 3.79 ms

b. To determine the speed of the cart before the loop-the-loop we apply conservation of energy

between the top of the ramp and the bottom. Thus we have

ΔE = ΔK + ΔUg + ΔUs = 0

0 = 12 mv f

2 − 12 mvi

2( ) + mgy f − mgyi( ) + 12 kx f

2 − 12 kxi

2( )0 = 1

2 mvbottom2 − 1

2 mvtop2( ) + mgybottom − mgytop( )

→ vbottom = 2g ytop − ybottom( ) + vtop2 = 2× 9.8 m

s2 ×15m+ 3.79 ms( )2

= 17.56 ms

c. Again apply conservation of energy between the bottom and the top of the loop-the-loop

and we have

ΔE = ΔK + ΔUg + ΔUs = 0

0 = 12 mv f

2 − 12 mvi

2( ) + mgy f − mgyi( ) + 12 kx f

2 − 12 kxi

2( )0 = 1

2 mvtop2 − 1

2 mvbottom2( ) + mgytop − mgybottom( )

ytop − ybottom = D =vbottom

2 − vtop2

2g

D =vbottom

2 − vtop2

2g=

17.56 ms( )2

− 8.55 ms( )2

2× 9.8 ms2

= 12.0m

d. If we consider gravity to be an external force then the work done by gravity is the difference

in the kinetic energies of the cart and riders between the top and bottom of the loop. WE have

Wg = ΔK = −ΔUg =

12 m vtop

2 − vbottom2( ) = −5.88×104 J = −58.8kJ

e. To calculate the new speed we consider the energy dissipated by friction. Noting that the

difference in energy is the energy dissipated as friction we have

ΔE fr =Wfr =12 mv f

2 − 12 mvi

2( ) + mgy f − mgyi( ) + 12 kx f

2 − 12 kxi

2( )ΔE fr =Wfr =

12 mv f

2 − 12 kxi

2

∴ 12 mv f

2 = 12 kxi

2 +Wfr → v f =km

xi2 +

2µK FN d cos 180( )m

=800 N

m

500kg−3.0m( )2

− 2× 0.2× 9.8 ms2 × 3m

v f = 1.62 ms