# solutions - Shoreline School District

1.
Initial system energy
KE
Eint
Final system energy
During
PEgrav PEspring Wext
Q
KE
Eint
PEgrav PEspring
Bar chart graph key
KE
Kinetic energy
Eint
Internal energy
PEgrav
PEspring
Arbitrary height, same for both
(can be zero)
Wext
Q
Gravitational
potential energy
Spring
potential energy
Work done by
external forces
to system
Use g = 10 m/s2
for simplicity
2.
Initial system energy
KE
Eint
Final system energy
During
PEgrav PEspring Wext
Q
KE
Eint
PEgrav PEspring
Bar chart graph key
KE
Kinetic energy
Eint
Internal energy
PEgrav
Arbitrary height, same for both
(can be zero)
PEspring
Wext
Q
Gravitational
potential energy
Spring
potential energy
Work done by
external forces
to system
Use g = 10 m/s2
for simplicity
3.
Answer: The bar given in the problem represents 320 Joules of work (8 bars); the final spring potential
energy is 160 Joules. The work done by friction is negative 80 Joules (2 bars), and there must therefore be
320 - 80 -160 = 80 Joules of kinetic energy.
Initial system energy
KE
Eint
Final system energy
During
PEgrav PEspring Wext
Q
KE
Eint
PEgrav PEspring
Bar chart graph key
KE
Kinetic energy
Eint
Internal energy
PEgrav
PEspring
80 Joules more
than initial
Wext
Arbitrary initial height
(can be zero)
Q
Gravitational
potential energy
Spring
potential energy
Work done by
external forces
to system
Use g = 10 m/s2
for simplicity
4. Answer: note that some energy is changed into internal energy of the system such as heat (spring)
Initial system energy
KE
Eint
Final system energy
During
PEgrav PEspring Wext
Q
KE
Eint
PEgrav PEspring
Bar chart graph key
KE
Kinetic energy
Eint
Internal energy
PEgrav
Arbitrary initial height
(can be zero)
PEspring
Wext
Q
Gravitational
potential energy
Spring
potential energy
Work done by
external forces
to system
Use g = 10 m/s2
for simplicity